Chapter 8 Binomial Theorem

Given:

The given binomial expansion is $\left( x + \frac{1}{x} \right)^{2r}$.

To Find:

The r^th term in the expansion.

Solution:

The general term, $T_{k+1}$, in the binomial expansion of $(a+b)^n$ is given by the formula:

$T_{k+1} = \binom{n}{k} a^{n-k} b^k$

For the given expansion $\left( x + \frac{1}{x} \right)^{2r}$, we compare it with $(a+b)^n$ to identify the values of $a$, $b$, and $n$:

$a = x$

$b = \frac{1}{x}$

$n = 2r$

We are asked to find the r^th term of the expansion. The r^th term is denoted by $T_r$.

Using the general term formula $T_{k+1} = \binom{n}{k} a^{n-k} b^k$, we set $k+1$ equal to the term number we want to find, which is $r$.

$k+1 = r$

Solving for $k$, we get:

$k = r-1$

Now, substitute the values of $n$, $k$, $a$, and $b$ into the general term formula:

$T_r = T_{(r-1)+1} = \binom{2r}{r-1} (x)^{2r - (r-1)} \left( \frac{1}{x} \right)^{r-1}$

Simplify the exponents of $x$ and $\frac{1}{x}$:

$T_r = \binom{2r}{r-1} x^{2r - r + 1} (x^{-1})^{r-1}$

$T_r = \binom{2r}{r-1} x^{r + 1} x^{-(r-1)}$

$T_r = \binom{2r}{r-1} x^{r + 1} x^{-r + 1}$

Using the property of exponents $x^m \cdot x^p = x^{m+p}$, combine the terms with $x$:

$T_r = \binom{2r}{r-1} x^{(r+1) + (-r+1)}$

$T_r = \binom{2r}{r-1} x^{r+1-r+1}$

$T_r = \binom{2r}{r-1} x^2$

Therefore, the r^th term in the expansion of $\left( x + \frac{1}{x} \right)^{2r}$ is $\binom{2r}{r-1} x^2$.

The final answer is $\boxed{\binom{2r}{r-1} x^2}$.

Given:

The expression to be expanded is $(1 - x + x^2)^4$.

To Expand:

We need to find the complete expansion of the given expression.

Solution (using Multinomial Theorem):

The multinomial theorem states that the expansion of $(a_1 + a_2 + \dots + a_m)^n$ is given by:

$\sum_{n_1+n_2+\dots+n_m=n} \frac{n!}{n_1! n_2! \dots n_m!} a_1^{n_1} a_2^{n_2} \dots a_m^{n_m}$

In this case, we have a trinomial $(1 + (-x) + x^2)^4$. So, $m=3$, $n=4$, $a_1=1$, $a_2=-x$, and $a_3=x^2$.

The general term in the expansion is $\frac{4!}{n_1! n_2! n_3!} (1)^{n_1} (-x)^{n_2} (x^2)^{n_3}$, where $n_1, n_2, n_3$ are non-negative integers such that $n_1 + n_2 + n_3 = 4$.

The general term simplifies to:

$T = \frac{4!}{n_1! n_2! n_3!} (1) (-1)^{n_2} x^{n_2} x^{2n_3} = \frac{4!}{n_1! n_2! n_3!} (-1)^{n_2} x^{n_2 + 2n_3}$

We need to find all combinations of non-negative integers $(n_1, n_2, n_3)$ such that $n_1 + n_2 + n_3 = 4$ and calculate the corresponding terms:

Now, we sum the terms with the same power of $x$:

$x^0$: $1$

$x^1$: $-4x$

$x^2$: $4x^2 + 6x^2 = 10x^2$

$x^3$: $-12x^3 - 4x^3 = -16x^3$

$x^4$: $6x^4 + 12x^4 + x^4 = 19x^4$

$x^5$: $-12x^5 - 4x^5 = -16x^5$

$x^6$: $4x^6 + 6x^6 = 10x^6$

$x^7$: $-4x^7$

$x^8$: $x^8$

Combining these terms, the expansion is:

$1 - 4x + 10x^2 - 16x^3 + 19x^4 - 16x^5 + 10x^6 - 4x^7 + x^8$

Alternate Solution (by squaring twice):

We can write $(1 - x + x^2)^4$ as $((1 - x + x^2)^2)^2$.

First, let's expand $(1 - x + x^2)^2$:

$(1 - x + x^2)^2 = (1 - (x - x^2))^2 = 1^2 - 2(1)(x - x^2) + (x - x^2)^2$

$= 1 - 2x + 2x^2 + (x^2 - 2(x)(x^2) + (x^2)^2)$

$= 1 - 2x + 2x^2 + x^2 - 2x^3 + x^4$

$= 1 - 2x + 3x^2 - 2x^3 + x^4$

Now, we need to square this result:

$(1 - 2x + 3x^2 - 2x^3 + x^4)^2$

We can expand this as the square of a polynomial:

$(a+b+c+d+e)^2 = a^2+b^2+c^2+d^2+e^2 + 2ab+2ac+2ad+2ae+2bc+2bd+2be+2cd+2ce+2de$

Let $a=1, b=-2x, c=3x^2, d=-2x^3, e=x^4$.

Squares of terms:

$a^2 = (1)^2 = 1$

$b^2 = (-2x)^2 = 4x^2$

$c^2 = (3x^2)^2 = 9x^4$

$d^2 = (-2x^3)^2 = 4x^6$

$e^2 = (x^4)^2 = x^8$

Twice the products of pairs of terms:

$2ab = 2(1)(-2x) = -4x$

$2ac = 2(1)(3x^2) = 6x^2$

$2ad = 2(1)(-2x^3) = -4x^3$

$2ae = 2(1)(x^4) = 2x^4$

$2bc = 2(-2x)(3x^2) = -12x^3$

$2bd = 2(-2x)(-2x^3) = 8x^4$

$2be = 2(-2x)(x^4) = -4x^5$

$2cd = 2(3x^2)(-2x^3) = -12x^5$

$2ce = 2(3x^2)(x^4) = 6x^6$

$2de = 2(-2x^3)(x^4) = -4x^7$

Summing all these terms and collecting by powers of $x$:

$x^0$: $1$

$x^1$: $-4x$

$x^2$: $4x^2 + 6x^2 = 10x^2$

$x^3$: $-4x^3 - 12x^3 = -16x^3$

$x^4$: $9x^4 + 2x^4 + 8x^4 = 19x^4$

$x^5$: $-4x^5 - 12x^5 = -16x^5$

$x^6$: $4x^6 + 6x^6 = 10x^6$

$x^7$: $-4x^7$

$x^8$: $x^8$

Combining these terms, the expansion is:

$1 - 4x + 10x^2 - 16x^3 + 19x^4 - 16x^5 + 10x^6 - 4x^7 + x^8$

Both methods yield the same result.

The final expanded form of $(1 - x + x^2)^4$ is $1 - 4x + 10x^2 - 16x^3 + 19x^4 - 16x^5 + 10x^6 - 4x^7 + x^8$.

The final answer is $\boxed{1 - 4x + 10x^2 - 16x^3 + 19x^4 - 16x^5 + 10x^6 - 4x^7 + x^8}$.

Given:

The binomial expansion $\left( \frac{x^3}{2} − \frac{2}{x^2} \right)^9$.

To Find:

The 4^th term from the end of the expansion.

Solution:

The given expansion is of the form $(a+b)^n$, where $a = \frac{x^3}{2}$, $b = -\frac{2}{x^2}$, and $n = 9$.

The total number of terms in the expansion of $(a+b)^n$ is $n+1$.

In this case, the total number of terms is $9+1 = 10$.

To find the r^th term from the end of an expansion with $N$ terms, we find the $(N - r + 1)^{\text{th}}$ term from the beginning.

Here, $N=10$ and $r=4$.

The 4^th term from the end is the $(10 - 4 + 1)^{\text{th}}$ term from the beginning.

This is the $(6+1)^{\text{th}}$ term from the beginning, which is the 7^th term from the beginning.

The general term $T_{k+1}$ in the expansion of $(a+b)^n$ is given by:

$T_{k+1} = \binom{n}{k} a^{n-k} b^k$

We need to find the 7^th term from the beginning, so we set $k+1 = 7$, which means $k=6$.

Substitute the values $n=9$, $k=6$, $a = \frac{x^3}{2}$, and $b = -\frac{2}{x^2}$ into the general term formula:

$T_7 = \binom{9}{6} \left(\frac{x^3}{2}\right)^{9-6} \left(-\frac{2}{x^2}\right)^6$

$T_7 = \binom{9}{6} \left(\frac{x^3}{2}\right)^3 \left(-\frac{2}{x^2}\right)^6$

Calculate the binomial coefficient $\binom{9}{6}$:

$\binom{9}{6} = \binom{9}{9-6} = \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7 \times \cancel{6!}}{(3 \times 2 \times 1) \times \cancel{6!}} = \frac{9 \times 8 \times 7}{6} = 3 \times 4 \times 7 = 84$

Calculate the powers of the terms:

$\left(\frac{x^3}{2}\right)^3 = \frac{(x^3)^3}{2^3} = \frac{x^9}{8}$

$\left(-\frac{2}{x^2}\right)^6 = (-1)^6 \left(\frac{2}{x^2}\right)^6 = 1 \cdot \frac{2^6}{(x^2)^6} = \frac{64}{x^{12}}$

Now substitute these values back into the expression for $T_7$:

$T_7 = 84 \cdot \frac{x^9}{8} \cdot \frac{64}{x^{12}}$

Simplify the numerical and variable parts:

$T_7 = 84 \cdot \frac{\cancel{64}^8}{\cancel{8}_1} \cdot \frac{x^9}{x^{12}}$

$T_7 = 84 \cdot 8 \cdot x^{9-12}$

$T_7 = 672 \cdot x^{-3}$

$T_7 = \frac{672}{x^3}$

Alternate Solution:

The r^th term from the end in the expansion of $(a+b)^n$ is the same as the r^th term from the beginning in the expansion of $(b+a)^n$.

So, the 4^th term from the end in $\left( \frac{x^3}{2} − \frac{2}{x^2} \right)^9$ is the 4^th term from the beginning in $\left( -\frac{2}{x^2} + \frac{x^3}{2} \right)^9$.

For the expansion $\left( -\frac{2}{x^2} + \frac{x^3}{2} \right)^9$, we have $a' = -\frac{2}{x^2}$, $b' = \frac{x^3}{2}$, and $n=9$.

We need the 4^th term from the beginning, $T_4$. For $T_4$, we set $k+1=4$, so $k=3$.

Using the general term formula $T_{k+1} = \binom{n}{k} (a')^{n-k} (b')^k$:

$T_4 = \binom{9}{3} \left(-\frac{2}{x^2}\right)^{9-3} \left(\frac{x^3}{2}\right)^3$

$T_4 = \binom{9}{3} \left(-\frac{2}{x^2}\right)^6 \left(\frac{x^3}{2}\right)^3$

Calculate the terms:

$\binom{9}{3} = 84$ (as calculated before)

$\left(-\frac{2}{x^2}\right)^6 = (-1)^6 \left(\frac{2}{x^2}\right)^6 = 1 \cdot \frac{2^6}{(x^2)^6} = \frac{64}{x^{12}}$

$\left(\frac{x^3}{2}\right)^3 = \frac{(x^3)^3}{2^3} = \frac{x^9}{8}$

Substitute these values back into the expression for $T_4$:

$T_4 = 84 \cdot \frac{64}{x^{12}} \cdot \frac{x^9}{8}$

$T_4 = 84 \cdot \frac{\cancel{64}^8}{\cancel{8}_1} \cdot \frac{x^9}{x^{12}}$

$T_4 = 84 \cdot 8 \cdot x^{9-12}$

$T_4 = 672 \cdot x^{-3}$

$T_4 = \frac{672}{x^3}$

Both methods confirm that the 4^th term from the end is $\frac{672}{x^3}$.

The final answer is $\boxed{\frac{672}{x^3}}$.

Given:

The expression to be evaluated is $\left( x^2 − \sqrt{1 − x^2} \right)^4 + \left( x^2 + \sqrt{1 − x^2} \right)^4$.

To Evaluate:

We need to simplify and find the value of the given expression.

Solution:

The given expression is in the form $(a-b)^n + (a+b)^n$, where $a = x^2$, $b = \sqrt{1-x^2}$, and $n=4$.

We know the binomial expansions for $(a+b)^n$ and $(a-b)^n$:

$(a+b)^n = \binom{n}{0} a^n b^0 + \binom{n}{1} a^{n-1} b^1 + \binom{n}{2} a^{n-2} b^2 + \binom{n}{3} a^{n-3} b^3 + \dots + \binom{n}{n} a^0 b^n$

$(a-b)^n = \binom{n}{0} a^n b^0 - \binom{n}{1} a^{n-1} b^1 + \binom{n}{2} a^{n-2} b^2 - \binom{n}{3} a^{n-3} b^3 + \dots + (-1)^n \binom{n}{n} a^0 b^n$

Adding these two expansions, the terms with odd powers of $b$ cancel out:

$(a+b)^n + (a-b)^n = 2 \left[ \binom{n}{0} a^n b^0 + \binom{n}{2} a^{n-2} b^2 + \binom{n}{4} a^{n-4} b^4 + \dots \right]$

In this case, $n=4$, $a=x^2$, and $b=\sqrt{1-x^2}$. The sum will include terms with even powers of $b$ up to $b^4$. The relevant values of $k$ are $0, 2, 4$.

So, the expression becomes:

$2 \left[ \binom{4}{0} (x^2)^{4-0} (\sqrt{1-x^2})^0 + \binom{4}{2} (x^2)^{4-2} (\sqrt{1-x^2})^2 + \binom{4}{4} (x^2)^{4-4} (\sqrt{1-x^2})^4 \right]$

Calculate the binomial coefficients:

$\binom{4}{0} = 1$

$\binom{4}{2} = \frac{4!}{2!2!} = \frac{24}{4} = 6$

$\binom{4}{4} = 1$

Calculate the powers of $a$ and $b$:

$(x^2)^{4} = x^8$

$(x^2)^{2} = x^4$

$(x^2)^{0} = 1$

$(\sqrt{1-x^2})^0 = 1$

$(\sqrt{1-x^2})^2 = 1-x^2$

$(\sqrt{1-x^2})^4 = ((1-x^2)^{1/2})^4 = (1-x^2)^2 = 1 - 2x^2 + x^4$

Substitute these values back into the expression:

$2 \left[ 1 \cdot x^8 \cdot 1 + 6 \cdot x^4 \cdot (1-x^2) + 1 \cdot 1 \cdot (1 - 2x^2 + x^4) \right]$

$= 2 \left[ x^8 + 6x^4 - 6x^6 + 1 - 2x^2 + x^4 \right]$

Combine like terms:

$= 2 \left[ x^8 - 6x^6 + (6x^4 + x^4) - 2x^2 + 1 \right]$

$= 2 \left[ x^8 - 6x^6 + 7x^4 - 2x^2 + 1 \right]$

Distribute the factor of 2:

$= 2x^8 - 12x^6 + 14x^4 - 4x^2 + 2$

The final evaluated expression is $2x^8 - 12x^6 + 14x^4 - 4x^2 + 2$.

The final answer is $\boxed{2x^8 - 12x^6 + 14x^4 - 4x^2 + 2}$.

Given:

The binomial expansion is $\left( x^3 - \frac{2}{x^2} \right)^{12}$.

To Find:

The coefficient of $x^{11}$ in the given expansion.

Solution:

The given expansion is of the form $(a+b)^n$, where $a = x^3$, $b = -\frac{2}{x^2}$, and $n = 12$.

The general term, $T_{k+1}$, in the expansion of $(a+b)^n$ is given by the formula:

$T_{k+1} = \binom{n}{k} a^{n-k} b^k$

Substitute the values of $n$, $a$, and $b$ for the given expansion:

$T_{k+1} = \binom{12}{k} (x^3)^{12-k} \left(-\frac{2}{x^2}\right)^k$

Now, let's simplify the terms involving $x$ to find the power of $x$ in the general term:

$(x^3)^{12-k} = x^{3(12-k)} = x^{36-3k}$

$\left(-\frac{2}{x^2}\right)^k = (-2)^k \left(\frac{1}{x^2}\right)^k = (-2)^k (x^{-2})^k = (-2)^k x^{-2k}$

The variable part of the general term is the product of the terms involving $x$:

$x^{36-3k} \cdot x^{-2k} = x^{(36-3k) + (-2k)} = x^{36-5k}$

We want to find the coefficient of $x^{11}$. Therefore, we set the power of $x$ in the general term equal to 11:

$36 - 5k = 11$

Subtract 36 from both sides:

$-5k = 11 - 36$

$-5k = -25$

Divide by -5:

$k = \frac{-25}{-5}$

$k = 5$

Since $k=5$ is a non-negative integer and $0 \leq 5 \leq 12$, this value of $k$ is valid for the expansion.

The term containing $x^{11}$ corresponds to $k=5$. The coefficient of this term is the part of $T_{k+1}$ that does not involve $x$, which is $\binom{12}{k} (-2)^k$ for $k=5$.

Coefficient of $x^{11} = \binom{12}{5} (-2)^5$

Calculate $\binom{12}{5}$:

$\binom{12}{5} = \frac{12!}{5!(12-5)!} = \frac{12!}{5!7!} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times \cancel{7!}}{(5 \times 4 \times 3 \times 2 \times 1) \times \cancel{7!}} = \frac{12 \times 11 \times 10 \times 9 \times 8}{120}$

$\binom{12}{5} = \frac{\cancel{120}^{1} \times 11 \times \cancel{10}^{1} \times 9 \times 8}{\cancel{120}_{1}} = 11 \times 9 \times 8 = 11 \times 72 = 792$

Calculate $(-2)^5$:

$(-2)^5 = -32$

Now, multiply the results:

Coefficient of $x^{11} = 792 \times (-32)$

$792 \times (-32) = -25344$

The coefficient of $x^{11}$ in the expansion of $\left( x^3 - \frac{2}{x^2} \right)^{12}$ is $-25344$.

The final answer is $\boxed{-25344}$.

Given:

The binomial expansion is $\left( x^2 - \frac{2}{x} \right)^{18}$.

To Determine:

Whether the expansion contains a term with $x^{10}$.

Solution:

The given expansion is of the form $(a+b)^n$, where $a = x^2$, $b = -\frac{2}{x}$, and $n = 18$.

The general term, $T_{k+1}$, in the expansion of $(a+b)^n$ is given by the formula:

$T_{k+1} = \binom{n}{k} a^{n-k} b^k$

Substitute the values of $n$, $a$, and $b$ for the given expansion:

$T_{k+1} = \binom{18}{k} (x^2)^{18-k} \left(-\frac{2}{x}\right)^k$

Simplify the terms involving $x$ to find the power of $x$ in the general term:

$(x^2)^{18-k} = x^{2(18-k)} = x^{36-2k}$

$\left(-\frac{2}{x}\right)^k = (-2)^k \left(\frac{1}{x}\right)^k = (-2)^k (x^{-1})^k = (-2)^k x^{-k}$

The variable part of the general term is the product of the terms involving $x$:

$x^{36-2k} \cdot x^{-k} = x^{(36-2k) - k} = x^{36-3k}$

To determine if there is a term containing $x^{10}$, we set the power of $x$ in the general term equal to 10:

$36 - 3k = 10$

Now, we solve this equation for $k$:

$36 - 10 = 3k$

$26 = 3k$

$k = \frac{26}{3}$

The index $k$ in the general term $T_{k+1}$ must be a non-negative integer, where $0 \leq k \leq n$. In this case, $n=18$, so $k$ must be an integer such that $0 \leq k \leq 18$.

Since the calculated value of $k = \frac{26}{3}$ is not an integer, there is no integer value of $k$ that will result in a term with $x^{10}$.

Therefore, the expansion of $\left( x^2 - \frac{2}{x} \right)^{18}$ does not contain a term with $x^{10}$.

The final answer is $\boxed{\text{No}}$.

Given:

The binomial expansion is $\left( \frac{\sqrt{x}}{\sqrt{3}} + \frac{\sqrt{3}}{2x^3} \right)^{10}$.

To Find:

The term independent of $x$ in the expansion.

Solution:

The given expansion is of the form $(a+b)^n$, where $a = \frac{\sqrt{x}}{\sqrt{3}}$, $b = \frac{\sqrt{3}}{2x^3}$, and $n = 10$.

We can write $a$ and $b$ using exponents:

$a = \frac{x^{1/2}}{3^{1/2}} = \frac{1}{\sqrt{3}} x^{1/2}$

$b = \frac{3^{1/2}}{2 x^3} = \frac{\sqrt{3}}{2} x^{-3}$

The general term, $T_{k+1}$, in the expansion of $(a+b)^n$ is given by the formula:

$T_{k+1} = \binom{n}{k} a^{n-k} b^k$

Substitute the values of $n$, $a$, and $b$ for the given expansion:

$T_{k+1} = \binom{10}{k} \left(\frac{1}{\sqrt{3}} x^{1/2}\right)^{10-k} \left(\frac{\sqrt{3}}{2} x^{-3}\right)^k$

Now, let's simplify the terms involving $x$ and the constant terms separately.

The terms involving $x$ are:

$(x^{1/2})^{10-k} \cdot (x^{-3})^k = x^{(1/2)(10-k)} \cdot x^{-3k} = x^{\frac{10-k}{2}} \cdot x^{-3k}$

Combine the powers of $x$:

$x^{\frac{10-k}{2} - 3k} = x^{\frac{10-k - 6k}{2}} = x^{\frac{10-7k}{2}}$

The constant terms are:

$\binom{10}{k} \left(\frac{1}{\sqrt{3}}\right)^{10-k} \left(\frac{\sqrt{3}}{2}\right)^k = \binom{10}{k} (3^{-1/2})^{10-k} \frac{(3^{1/2})^k}{2^k}$

$= \binom{10}{k} 3^{-(1/2)(10-k)} \frac{3^{k/2}}{2^k} = \binom{10}{k} 3^{-5 + k/2} \frac{3^{k/2}}{2^k} = \binom{10}{k} \frac{3^{k/2 + k/2}}{3^5 \cdot 2^k} = \binom{10}{k} \frac{3^k}{3^5 \cdot 2^k} = \binom{10}{k} \frac{1}{3^{5-k} 2^k}$

The general term is:

$T_{k+1} = \binom{10}{k} \frac{3^{k-5}}{2^k} x^{\frac{10-7k}{2}}$

For the term independent of $x$, the power of $x$ must be zero. So, we set the exponent of $x$ equal to 0:

$\frac{10-7k}{2} = 0$

Multiply both sides by 2:

$10 - 7k = 0$

Solve for $k$:

$10 = 7k$

$k = \frac{10}{7}$

The index $k$ in the binomial expansion must be a non-negative integer such that $0 \leq k \leq n$. In this case, $n=10$, so $k$ must be an integer from 0 to 10.

Since the calculated value of $k = \frac{10}{7}$ is not an integer, there is no integer value of $k$ for which the power of $x$ is 0.

Therefore, there is no term independent of $x$ in the expansion of $\left( \frac{\sqrt{x}}{\sqrt{3}} + \frac{\sqrt{3}}{2x^3} \right)^{10}$.

The expansion does not contain a term independent of $x$.

The final answer is $\boxed{\text{No term independent of } x}$.

Given:

The binomial expansion is $\left( 2ax - \frac{b}{x^2} \right)^{12}$.

To Find:

The middle term in the expansion.

Solution:

The given expansion is of the form $(a+b)^n$, where $a = 2ax$, $b = -\frac{b}{x^2}$, and $n = 12$.

The total number of terms in the expansion of $(a+b)^n$ is $n+1$.

In this case, $n=12$, so the total number of terms is $12+1 = 13$.

Since the number of terms (13) is odd, there is only one middle term.

The position of the middle term is $\left( \frac{n}{2} + 1 \right)^{\text{th}}$ term when $n$ is even.

Here, $n=12$, so the middle term is the $\left( \frac{12}{2} + 1 \right)^{\text{th}} = (6+1)^{\text{th}} = 7^{\text{th}}$ term.

We need to find the 7^th term, $T_7$, in the expansion.

The general term, $T_{k+1}$, in the expansion of $(a+b)^n$ is given by the formula:

$T_{k+1} = \binom{n}{k} a^{n-k} b^k$

For the 7^th term, we have $k+1 = 7$, so $k=6$.

Substitute the values of $n=12$, $k=6$, $a=2ax$, and $b=-\frac{b}{x^2}$ into the general term formula:

$T_7 = \binom{12}{6} (2ax)^{12-6} \left(-\frac{b}{x^2}\right)^6$

$T_7 = \binom{12}{6} (2ax)^6 \left(-\frac{b}{x^2}\right)^6$

Calculate the binomial coefficient $\binom{12}{6}$:

$\binom{12}{6} = \frac{12!}{6!6!} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times \cancel{6!}}{(\cancel{6 \times 5 \times 4 \times 3 \times 2 \times 1}) \times \cancel{6!}} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{720}$

$\binom{12}{6} = \frac{\cancel{12}^{1} \times 11 \times \cancel{10}^{1} \times \cancel{9}^{3} \times \cancel{8}^{2} \times 7}{\cancel{720}_{60}} = \frac{11 \times 1 \times 3 \times 2 \times 7}{60} = \frac{462}{60}$ (Let's recalculate using simplification before multiplying)

$\binom{12}{6} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = \frac{\cancel{12}^{2} \times 11 \times \cancel{10}^{2} \times \cancel{9}^{3} \times \cancel{8}^{1} \times 7}{\cancel{6 \times 5 \times 4 \times 3 \times 2 \times 1}_{1 \times 1 \times 1 \times 1 \times 1 \times 1}}$ (Simplifying carefully)

$\binom{12}{6} = \frac{\cancel{12}^{1} \times 11 \times \cancel{10}^{1} \times \cancel{9}^{1} \times \cancel{8}^{1} \times 7}{\cancel{6}_{1} \times \cancel{5}_{1} \times \cancel{4}_{1} \times \cancel{3}_{1} \times \cancel{2}_{1} \times \cancel{1}_{1}}$ (Trying again with better cancellation visualization)

$\binom{12}{6} = \frac{\cancel{12}^{2} \times 11 \times \cancel{10}^{1} \times \cancel{9}^{1} \times 8 \times 7}{\cancel{6}_{1} \times 5 \times \cancel{4}_{1} \times \cancel{3}_{1} \times \cancel{2}_{1} \times 1} = \frac{2 \times 11 \times 1 \times 1 \times 8 \times 7}{5}$ (Still not right)

Let's do it step by step: $6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$.

$12 \times 11 \times 10 \times 9 \times 8 \times 7 = 665280$.

$\binom{12}{6} = \frac{665280}{720} = 924$.

Calculate the powers of the terms:

$(2ax)^6 = 2^6 a^6 x^6 = 64 a^6 x^6$

$\left(-\frac{b}{x^2}\right)^6 = (-1)^6 \frac{b^6}{(x^2)^6} = 1 \cdot \frac{b^6}{x^{12}} = \frac{b^6}{x^{12}}$

Substitute these values back into the expression for $T_7$:

$T_7 = 924 \cdot (64 a^6 x^6) \cdot \left(\frac{b^6}{x^{12}}\right)$

$T_7 = 924 \cdot 64 \cdot a^6 \cdot b^6 \cdot \frac{x^6}{x^{12}}$

Simplify the numerical and variable parts:

$924 \times 64 = 59136$

$\frac{x^6}{x^{12}} = x^{6-12} = x^{-6} = \frac{1}{x^6}$

$T_7 = 59136 a^6 b^6 \frac{1}{x^6}$

$T_7 = \frac{59136 a^6 b^6}{x^6}$

The middle term in the expansion is $\frac{59136 a^6 b^6}{x^6}$.

The final answer is $\boxed{\frac{59136 a^6 b^6}{x^6}}$.

Given:

The binomial expansion is $\left( \frac{p}{x} + \frac{x}{p} \right)^9$.

To Find:

The middle term(s) in the expansion.

Solution:

The given expansion is of the form $(a+b)^n$, where $a = \frac{p}{x}$, $b = \frac{x}{p}$, and $n = 9$.

The total number of terms in the expansion of $(a+b)^n$ is $n+1$.

In this case, $n=9$, so the total number of terms is $9+1 = 10$.

Since the total number of terms (10) is even, there are two middle terms.

The positions of the middle terms are the $\left(\frac{10}{2}\right)^{\text{th}}$ and $\left(\frac{10}{2} + 1\right)^{\text{th}}$ terms.

These are the 5^th term and the 6^th term.

The general term, $T_{k+1}$, in the expansion of $(a+b)^n$ is given by the formula:

$T_{k+1} = \binom{n}{k} a^{n-k} b^k$

Finding the 5^th term:

For the 5^th term, we have $k+1 = 5$, so $k=4$.

Substitute $n=9$, $k=4$, $a=\frac{p}{x}$, and $b=\frac{x}{p}$ into the general term formula:

$T_5 = \binom{9}{4} \left(\frac{p}{x}\right)^{9-4} \left(\frac{x}{p}\right)^4$

$T_5 = \binom{9}{4} \left(\frac{p}{x}\right)^5 \left(\frac{x}{p}\right)^4$

Calculate the binomial coefficient $\binom{9}{4}$:

$\binom{9}{4} = \frac{9!}{4!(9-4)!} = \frac{9!}{4!5!} = \frac{9 \times 8 \times 7 \times 6 \times \cancel{5!}}{(4 \times 3 \times 2 \times 1) \times \cancel{5!}} = \frac{9 \times 8 \times 7 \times 6}{24}$

$\binom{9}{4} = \frac{\cancel{9}^{3} \times \cancel{8}^{1} \times 7 \times \cancel{6}^{1}}{\cancel{24}_{1}} = 3 \times 1 \times 7 \times 1 = 126$

Calculate the powers of the terms:

$\left(\frac{p}{x}\right)^5 = \frac{p^5}{x^5}$

$\left(\frac{x}{p}\right)^4 = \frac{x^4}{p^4}$

Substitute these values back into the expression for $T_5$:

$T_5 = 126 \cdot \frac{p^5}{x^5} \cdot \frac{x^4}{p^4}$

$T_5 = 126 \cdot p^{5-4} \cdot x^{4-5}$

$T_5 = 126 p^1 x^{-1} = \frac{126p}{x}$

Finding the 6^th term:

For the 6^th term, we have $k+1 = 6$, so $k=5$.

Substitute $n=9$, $k=5$, $a=\frac{p}{x}$, and $b=\frac{x}{p}$ into the general term formula:

$T_6 = \binom{9}{5} \left(\frac{p}{x}\right)^{9-5} \left(\frac{x}{p}\right)^5$

$T_6 = \binom{9}{5} \left(\frac{p}{x}\right)^4 \left(\frac{x}{p}\right)^5$

Calculate the binomial coefficient $\binom{9}{5}$:

$\binom{9}{5} = \binom{9}{9-5} = \binom{9}{4} = 126$ (as calculated for $T_5$)

Calculate the powers of the terms:

$\left(\frac{p}{x}\right)^4 = \frac{p^4}{x^4}$

$\left(\frac{x}{p}\right)^5 = \frac{x^5}{p^5}$

Substitute these values back into the expression for $T_6$:

$T_6 = 126 \cdot \frac{p^4}{x^4} \cdot \frac{x^5}{p^5}$

$T_6 = 126 \cdot p^{4-5} \cdot x^{5-4}$

$T_6 = 126 p^{-1} x^1 = \frac{126x}{p}$

The middle terms in the expansion are the 5^th term and the 6^th term, which are $\frac{126p}{x}$ and $\frac{126x}{p}$.

The final answer is $\boxed{\frac{126p}{x}, \frac{126x}{p}}$.

Given:

The expression is $2^{4n + 4} – 15n – 16$, where $n \in \mathbb{N}$.

To Prove:

The expression $2^{4n + 4} – 15n – 16$ is divisible by 225 for all $n \in \mathbb{N}$.

Solution:

Let the given expression be $E$.

$E = 2^{4n + 4} – 15n – 16$

We can rewrite $2^{4n+4}$ as $2^{4(n+1)} = (2^4)^{n+1} = 16^{n+1}$.

So, $E = 16^{n+1} - 15n - 16$.

We know that $16 = 1 + 15$. Let's expand $16^{n+1}$ using the binomial theorem for $(1+x)^m$ with $x=15$ and $m=n+1$.

$(1+x)^m = \binom{m}{0} + \binom{m}{1}x + \binom{m}{2}x^2 + \binom{m}{3}x^3 + \dots + \binom{m}{m}x^m$

Substituting $m=n+1$ and $x=15$, we get:

$(1+15)^{n+1} = \binom{n+1}{0} + \binom{n+1}{1}(15) + \binom{n+1}{2}(15)^2 + \binom{n+1}{3}(15)^3 + \dots + \binom{n+1}{n+1}(15)^{n+1}$

We know that $\binom{n+1}{0} = 1$ and $\binom{n+1}{1} = n+1$.

So, the expansion becomes:

$16^{n+1} = 1 + (n+1)(15) + \binom{n+1}{2}(15)^2 + \binom{n+1}{3}(15)^3 + \dots + (15)^{n+1}$

$16^{n+1} = 1 + 15n + 15 + \binom{n+1}{2}(15^2) + \binom{n+1}{3}(15^3) + \dots + 15^{n+1}$

$16^{n+1} = 16 + 15n + \binom{n+1}{2}(15^2) + \binom{n+1}{3}(15^3) + \dots + 15^{n+1}$

Now, substitute this expansion back into the expression for $E$:

$E = (16 + 15n + \binom{n+1}{2}(15^2) + \binom{n+1}{3}(15^3) + \dots + 15^{n+1}) - 15n - 16$

Group the terms:

$E = (16 - 16) + (15n - 15n) + \binom{n+1}{2}(15^2) + \binom{n+1}{3}(15^3) + \dots + 15^{n+1}$

$E = 0 + 0 + \binom{n+1}{2}(15^2) + \binom{n+1}{3}(15^3) + \dots + 15^{n+1}$

$E = \binom{n+1}{2}(15^2) + \binom{n+1}{3}(15^3) + \dots + \binom{n+1}{n+1}(15)^{n+1}$

We know that $15^2 = 225$. Notice that every term from the first term onwards in this summation has a factor of $15^2$ or a higher power of 15.

We can factor out $15^2$ (which is 225) from each term:

$E = 15^2 \left[ \binom{n+1}{2} + \binom{n+1}{3}(15)^1 + \binom{n+1}{4}(15)^2 + \dots + \binom{n+1}{n+1}(15)^{n-1} \right]$

$E = 225 \left[ \binom{n+1}{2} + 15 \binom{n+1}{3} + 15^2 \binom{n+1}{4} + \dots + 15^{n-1} \binom{n+1}{n+1} \right]$

Let $K = \binom{n+1}{2} + 15 \binom{n+1}{3} + 15^2 \binom{n+1}{4} + \dots + 15^{n-1} \binom{n+1}{n+1}$.

For any natural number $n \in \mathbb{N}$, $n+1$ is an integer greater than or equal to 2.

The binomial coefficients $\binom{n+1}{k}$ are integers for all integers $k$ where $0 \leq k \leq n+1$.

The terms $15, 15^2, \dots, 15^{n-1}$ are also integers for $n \in \mathbb{N}$. (If $n=1$, the sum for $K$ starts from $k=2$ and ends at $k=n+1=2$, so $K = \binom{1+1}{2} = \binom{2}{2} = 1$, which is an integer. If $n \ge 2$, the powers of 15 are positive integers).

Therefore, the sum $K$ is an integer for all $n \in \mathbb{N}$.

So, $E = 225 \times K$, where $K$ is an integer.

This means that the expression $2^{4n+4} - 15n - 16$ is a multiple of 225 for all $n \in \mathbb{N}$.

Hence, $2^{4n + 4} – 15n – 16$ is divisible by 225 for all $n \in \mathbb{N}$.

The final answer is $\boxed{2^{4n + 4} – 15n – 16 \text{ is divisible by } 225}$.

Given:

Expansion: $(2 + 3x)^9$

Value of x: $x = \frac{3}{2}$

To Find:

Numerically greatest term.

Solution:

The expansion is $(2+3x)^9$. With $x=\frac{3}{2}$, the expression inside the parenthesis is $2 + 3(\frac{3}{2}) = 2 + \frac{9}{2}$.

We can rewrite the expansion as $2^9 \left( 1 + \frac{3x}{2} \right)^9$.

Let $y = \frac{3x}{2}$. Substituting $x=\frac{3}{2}$, $y = \frac{3}{2} \cdot \frac{3}{2} = \frac{9}{4}$.

We consider the expansion $(1+y)^n$, where $n=9$ and $y=\frac{9}{4}$.

The numerically greatest term $T_{r+1}$ is found by determining the integer $r$ such that $\left| \frac{T_{r+1}}{T_r} \right| \geq 1$.

$\left| \frac{n-r+1}{r} y \right| \geq 1$

Substituting $n=9$ and $y=\frac{9}{4}$:

$\left| \frac{9-r+1}{r} \cdot \frac{9}{4} \right| \geq 1$

$\left| \frac{10-r}{r} \cdot \frac{9}{4} \right| \geq 1$

Since $1 \leq r \leq 9$, we have $\frac{10-r}{r} \cdot \frac{9}{4} \geq 1$

$9(10-r) \geq 4r$

$90 - 9r \geq 4r$

$90 \geq 13r$

$r \leq \frac{90}{13} \approx 6.923$

The greatest integer value of $r$ satisfying this is $r=6$. This means $T_{r+1} = T_7$ is the numerically greatest term.

The general term in the expansion of $(a+b)^n$ is $T_{k+1} = \binom{n}{k} a^{n-k} b^k$.

For $(2+3x)^9$, $a=2$, $b=3x$, $n=9$. For the 7^th term, $k+1=7$, so $k=6$.

$T_7 = \binom{9}{6} (2)^{9-6} (3x)^6$

$T_7 = \binom{9}{6} (2)^3 (3x)^6$

Substitute $x=\frac{3}{2}$:

$T_7 = \binom{9}{6} (2)^3 \left(3 \cdot \frac{3}{2}\right)^6$

$T_7 = \binom{9}{6} (2)^3 \left(\frac{9}{2}\right)^6$

Calculate the terms:

$\binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$

$(2)^3 = 8$

$\left(\frac{9}{2}\right)^6 = \frac{9^6}{2^6} = \frac{531441}{64}$

$T_7 = 84 \cdot 8 \cdot \frac{531441}{64}$

$T_7 = 84 \cdot \frac{8}{64} \cdot 531441$

$T_7 = 84 \cdot \frac{1}{8} \cdot 531441$

$T_7 = \frac{84}{8} \cdot 531441$

$T_7 = 10.5 \cdot 531441$

$T_7 = 5580130.5$

The numerical value of the greatest term is $5580130.5$.

The final answer is $\boxed{5580130.5}$.

Given:

The expression is $(1 + x)^n \left( 1+\frac{1}{x} \right)^n$, where $n$ is a positive integer.

To Find:

The coefficient of $x^{-1}$ in the expansion of the given expression.

Solution:

Let the given expression be $E$.

$E = (1 + x)^n \left( 1+\frac{1}{x} \right)^n$

We can rewrite the second factor:

$1 + \frac{1}{x} = \frac{x+1}{x}$

Substitute this into the expression $E$:

$E = (1 + x)^n \left( \frac{x+1}{x} \right)^n$

Using the property $(ab)^m = a^m b^m$, we can write $\left( \frac{x+1}{x} \right)^n = \frac{(x+1)^n}{x^n}$.

$E = (1 + x)^n \cdot \frac{(1+x)^n}{x^n}$

Using the property $a^m a^p = a^{m+p}$, combine the terms with $(1+x)$:

$E = \frac{(1+x)^{n+n}}{x^n} = \frac{(1+x)^{2n}}{x^n}$

Rewrite $\frac{1}{x^n}$ as $x^{-n}$:

$E = x^{-n} (1+x)^{2n}$

Now, we expand $(1+x)^{2n}$ using the binomial theorem. The general term in the expansion of $(1+y)^m$ is $\binom{m}{k} y^k$. Here $y=x$ and $m=2n$.

$(1+x)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} x^k$

Substitute this expansion back into the expression for $E$:

$E = x^{-n} \sum_{k=0}^{2n} \binom{2n}{k} x^k$

Multiply $x^{-n}$ with each term in the sum:

$E = \sum_{k=0}^{2n} \binom{2n}{k} x^{-n} x^k$

Using the property $x^p x^q = x^{p+q}$:

$E = \sum_{k=0}^{2n} \binom{2n}{k} x^{k-n}$

We want to find the coefficient of $x^{-1}$ in this expansion. This means the power of $x$ must be equal to $-1$.

So, we set the exponent of $x$ equal to $-1$:

$k - n = -1$

Solve for $k$:

$k = n - 1$

Since $n$ is a positive integer ($n \in \mathbb{N}$), $n \geq 1$. The possible values for the index $k$ in the summation are integers from $0$ to $2n$. For $n \geq 1$, $k=n-1$ is an integer such that $0 \leq n-1 < 2n$, so $0 \leq k \leq 2n$ is satisfied.

The coefficient of $x^{-1}$ is the coefficient of the term where $k = n-1$. This coefficient is given by $\binom{2n}{k}$.

Coefficient of $x^{-1} = \binom{2n}{n-1}$

Using the symmetry property of binomial coefficients, $\binom{N}{K} = \binom{N}{N-K}$, we have:

$\binom{2n}{n-1} = \binom{2n}{2n - (n-1)} = \binom{2n}{2n - n + 1} = \binom{2n}{n+1}$

Both $\binom{2n}{n-1}$ and $\binom{2n}{n+1}$ are valid expressions for the coefficient.

The coefficient of $x^{-1}$ in the expansion of $(1 + x)^n \left( 1+\frac{1}{x} \right)^n$ is $\binom{2n}{n-1}$.

The final answer is $\boxed{\binom{2n}{n-1}}$.

Given:

Quantities to compare: $99^{50} + 100^{50}$ and $101^{50}$.

To Determine:

Which quantity is larger.

Solution:

We can write $101^{50} = (100+1)^{50}$ and $99^{50} = (100-1)^{50}$.

Using the binomial theorem:

$(a+b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots$

$(a-b)^n = \binom{n}{0}a^n - \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 - \dots$

Let $a=100$, $b=1$, and $n=50$.

$101^{50} = (100+1)^{50} = \binom{50}{0}100^{50} + \binom{50}{1}100^{49} + \binom{50}{2}100^{48} + \binom{50}{3}100^{47} + \dots$

$99^{50} = (100-1)^{50} = \binom{50}{0}100^{50} - \binom{50}{1}100^{49} + \binom{50}{2}100^{48} - \binom{50}{3}100^{47} + \dots$

Consider the difference between $101^{50}$ and $99^{50}$:

$101^{50} - 99^{50} = [(100+1)^{50}] - [(100-1)^{50}]$

$101^{50} - 99^{50} = \left[ \binom{50}{0}100^{50} + \binom{50}{1}100^{49} + \binom{50}{2}100^{48} + \dots \right] - \left[ \binom{50}{0}100^{50} - \binom{50}{1}100^{49} + \binom{50}{2}100^{48} - \dots \right]$

$101^{50} - 99^{50} = 2 \left[ \binom{50}{1}100^{49} + \binom{50}{3}100^{47} + \binom{50}{5}100^{45} + \dots + \binom{50}{49}100^1 \right]$

The first term inside the bracket is $\binom{50}{1}100^{49} = 50 \cdot 100^{49}$.

$101^{50} - 99^{50} = 2 \cdot (50 \cdot 100^{49}) + 2 \left[ \binom{50}{3}100^{47} + \dots \right]$

$101^{50} - 99^{50} = 100 \cdot 100^{49} + 2 \left[ \binom{50}{3}100^{47} + \dots \right]$

$101^{50} - 99^{50} = 100^{50} + 2 \left[ \binom{50}{3}100^{47} + \binom{50}{5}100^{45} + \dots + \binom{50}{49}100 \right]$

The terms $\binom{50}{3}, \binom{50}{5}, \dots, \binom{50}{49}$ are all positive integers, and the powers of 100 are also positive. Thus, the quantity inside the square brackets is positive.

$2 \left[ \binom{50}{3}100^{47} + \dots \right] > 0$

So, $101^{50} - 99^{50} = 100^{50} + (\text{a positive value})$

This implies that $101^{50} - 99^{50} > 100^{50}$.

Adding $99^{50}$ to both sides of the inequality:

$101^{50} > 100^{50} + 99^{50}$

Therefore, $101^{50}$ is larger.

The final answer is $\boxed{101^{50}}$.

Given:

The expression: $(1 + x)^{1000} + x (1 + x)^{999} + x^2 (1 + x)^{998} + \dots + x^{1000}$.

To Find:

The coefficient of $x^{50}$ in the simplified expansion.

Solution:

The given expression is a finite geometric series:

$S = (1 + x)^{1000} + x (1 + x)^{999} + x^2 (1 + x)^{998} + \dots + x^{1000}$

The first term is $A = (1 + x)^{1000}$.

The terms can be written as $x^k (1+x)^{1000-k}$ for $k=0, 1, 2, \dots, 1000$.

Term 0: $x^0 (1+x)^{1000-0} = (1+x)^{1000}$

Term 1: $x^1 (1+x)^{1000-1} = x(1+x)^{999}$

... environments.

Term 1000: $x^{1000} (1+x)^{1000-1000} = x^{1000}$

The common ratio is $R = \frac{x (1+x)^{999}}{(1+x)^{1000}} = \frac{x}{1+x}$.

The number of terms is $N = 1000 - 0 + 1 = 1001$.

The sum of a finite geometric series is $S = A \frac{1 - R^N}{1 - R}$.

$S = (1+x)^{1000} \frac{1 - \left(\frac{x}{1+x}\right)^{1001}}{1 - \frac{x}{1+x}}$

Simplify the denominator:

$1 - \frac{x}{1+x} = \frac{(1+x) - x}{1+x} = \frac{1}{1+x}$

Simplify the numerator of the fraction:

$1 - \left(\frac{x}{1+x}\right)^{1001} = 1 - \frac{x^{1001}}{(1+x)^{1001}} = \frac{(1+x)^{1001} - x^{1001}}{(1+x)^{1001}}$

Substitute back into the sum formula:

$S = (1+x)^{1000} \frac{\frac{(1+x)^{1001} - x^{1001}}{(1+x)^{1001}}}{\frac{1}{1+x}}$

$S = (1+x)^{1000} \cdot \frac{(1+x)^{1001} - x^{1001}}{(1+x)^{1001}} \cdot (1+x)$

$S = (1+x)^{1000} \cdot \frac{(1+x)^{1001} - x^{1001}}{(1+x)^{1000}}$

$S = (1+x)^{1001} - x^{1001}$

We need to find the coefficient of $x^{50}$ in the expansion of $(1+x)^{1001} - x^{1001}$.

Consider the two terms separately:

1. $(1+x)^{1001}$:

The binomial expansion is $(1+x)^{1001} = \sum_{k=0}^{1001} \binom{1001}{k} x^k$. The coefficient of $x^{50}$ in this term is obtained when $k=50$, which is $\binom{1001}{50}$.

2. $-x^{1001}$:

This term is $-x^{1001}$. The power of $x$ here is 1001. We are looking for the coefficient of $x^{50}$. Since $50 \neq 1001$, the coefficient of $x^{50}$ in this term is 0.

The coefficient of $x^{50}$ in the sum is the sum of the coefficients of $x^{50}$ from each term:

Coefficient of $x^{50}$ in $S = (\text{Coefficient of } x^{50} \text{ in } (1+x)^{1001}) + (\text{Coefficient of } x^{50} \text{ in } -x^{1001})$

Coefficient of $x^{50} = \binom{1001}{50} + 0 = \binom{1001}{50}$.

The coefficient of $x^{50}$ in the simplified expansion is $\binom{1001}{50}$.

The final answer is $\boxed{\binom{1001}{50}}$.

Given:

$a_1, a_2, a_3, a_4$ are the coefficients of four consecutive terms in the expansion of $(1+x)^n$, where $n$ is a positive integer.

To Prove:

$\frac{a_1}{a_1 + a_2} + \frac{a_3}{a_3 + a_4} = \frac{2a_2}{a_2 + a_3}$

Proof:

Let the four consecutive terms be the $(k+1)^{\text{th}}$, $(k+2)^{\text{th}}$, $(k+3)^{\text{th}}$, and $(k+4)^{\text{th}}$ terms in the expansion of $(1+x)^n$.

The general term in the expansion of $(1+x)^n$ is $T_{r+1} = \binom{n}{r} x^r$. The coefficient of the $(r+1)^{\text{th}}$ term is $\binom{n}{r}$.

So, the coefficients of the four consecutive terms are:

$a_1 = \binom{n}{k}$

$a_2 = \binom{n}{k+1}$

$a_3 = \binom{n}{k+2}$

$a_4 = \binom{n}{k+3}$

where $k$ is a non-negative integer such that $0 \leq k < k+1 < k+2 < k+3 \leq n$.

We use the property of binomial coefficients: $\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r}$.

Consider the ratios of consecutive coefficients:

$\frac{a_2}{a_1} = \frac{\binom{n}{k+1}}{\binom{n}{k}} = \frac{n-(k+1)+1}{k+1} = \frac{n-k}{k+1}$

$\frac{a_3}{a_2} = \frac{\binom{n}{k+2}}{\binom{n}{k+1}} = \frac{n-(k+2)+1}{k+2} = \frac{n-k-1}{k+2}$

$\frac{a_4}{a_3} = \frac{\binom{n}{k+3}}{\binom{n}{k+2}} = \frac{n-(k+3)+1}{k+3} = \frac{n-k-2}{k+3}$

Now, let's simplify the terms in the given identity:

Consider the term $\frac{a_1}{a_1 + a_2}$:

$\frac{a_1}{a_1 + a_2} = \frac{a_1}{a_1(1 + a_2/a_1)} = \frac{1}{1 + a_2/a_1}$

Substitute the ratio $\frac{a_2}{a_1}$:

$\frac{a_1}{a_1 + a_2} = \frac{1}{1 + \frac{n-k}{k+1}} = \frac{1}{\frac{k+1 + n-k}{k+1}} = \frac{k+1}{n+1}$

Consider the term $\frac{a_3}{a_3 + a_4}$:

$\frac{a_3}{a_3 + a_4} = \frac{a_3}{a_3(1 + a_4/a_3)} = \frac{1}{1 + a_4/a_3}$

Substitute the ratio $\frac{a_4}{a_3}$:

$\frac{a_3}{a_3 + a_4} = \frac{1}{1 + \frac{n-k-2}{k+3}} = \frac{1}{\frac{k+3 + n-k-2}{k+3}} = \frac{k+3}{n+1}$

Consider the term $\frac{a_2}{a_2 + a_3}$:

$\frac{a_2}{a_2 + a_3} = \frac{a_2}{a_2(1 + a_3/a_2)} = \frac{1}{1 + a_3/a_2}$

Substitute the ratio $\frac{a_3}{a_2}$:

$\frac{a_2}{a_2 + a_3} = \frac{1}{1 + \frac{n-k-1}{k+2}} = \frac{1}{\frac{k+2 + n-k-1}{k+2}} = \frac{k+2}{n+1}$

Now, substitute these simplified terms into the left side (LHS) of the identity:

LHS $= \frac{a_1}{a_1 + a_2} + \frac{a_3}{a_3 + a_4}$

LHS $= \frac{k+1}{n+1} + \frac{k+3}{n+1}$

LHS $= \frac{(k+1) + (k+3)}{n+1}$

LHS $= \frac{2k + 4}{n+1}$

Now, consider the right side (RHS) of the identity:

RHS $= \frac{2a_2}{a_2 + a_3}$

RHS $= 2 \cdot \left( \frac{a_2}{a_2 + a_3} \right)$

RHS $= 2 \cdot \frac{k+2}{n+1}$

RHS $= \frac{2(k+2)}{n+1} = \frac{2k + 4}{n+1}$

Comparing the LHS and RHS, we have:

LHS $= \frac{2k + 4}{n+1}$

RHS $= \frac{2k + 4}{n+1}$

Since LHS = RHS, the identity is proven.

Thus, if $a_1, a_2, a_3$ and $a_4$ are the coefficients of any four consecutive terms in the expansion of $(1+x)^n$, then $\frac{a_1}{a_1 + a_2} + \frac{a_3}{a_3 + a_4} = \frac{2a_2}{a_2 + a_3}$.

Given:

The expression $(x + a)^{51} – (x – a)^{51}$.

To Find:

The total number of terms after simplification.

Solution:

Let the given expression be $E$.

$E = (x + a)^{51} – (x – a)^{51}$

We use the binomial expansions for $(A+B)^n$ and $(A-B)^n$:

$(A+B)^n = \binom{n}{0}A^n B^0 + \binom{n}{1}A^{n-1}B^1 + \binom{n}{2}A^{n-2}B^2 + \dots + \binom{n}{n}A^0 B^n$

$(A-B)^n = \binom{n}{0}A^n B^0 - \binom{n}{1}A^{n-1}B^1 + \binom{n}{2}A^{n-2}B^2 - \dots + (-1)^n \binom{n}{n}A^0 B^n$

Subtracting the second expansion from the first:

$(A+B)^n - (A-B)^n = 2 \left[ \binom{n}{1}A^{n-1}B^1 + \binom{n}{3}A^{n-3}B^3 + \binom{n}{5}A^{n-5}B^5 + \dots \right]$

The terms that remain are those with odd indices in the binomial coefficient, multiplied by 2.

In our case, $A=x$, $B=a$, and $n=51$. So the expression is:

$(x+a)^{51} - (x-a)^{51} = 2 \left[ \binom{51}{1}x^{51-1}a^1 + \binom{51}{3}x^{51-3}a^3 + \binom{51}{5}x^{51-5}a^5 + \dots + \binom{51}{51}x^{51-51}a^{51} \right]$

$(x+a)^{51} - (x-a)^{51} = 2 \left[ \binom{51}{1}x^{50}a + \binom{51}{3}x^{48}a^3 + \binom{51}{5}x^{46}a^5 + \dots + \binom{51}{51}a^{51} \right]$

The powers of $a$ in the remaining terms are $1, 3, 5, \dots, 51$. Since these powers of $a$ are all distinct odd numbers, the terms $x^{51-k}a^k$ for $k \in \{1, 3, 5, \dots, 51\}$ are distinct terms and cannot be combined further by collecting like terms (assuming $x \neq 0$ and $a \neq 0$).

The number of terms in the simplified expansion is equal to the number of odd integers from 1 to 51.

To find the number of odd integers from 1 to 51, we can use the formula for an arithmetic sequence: Number of terms = $\frac{\text{Last term} - \text{First term}}{\text{Common difference}} + 1$.

The sequence of odd numbers is $1, 3, 5, \dots, 51$.

Number of terms = $\frac{51 - 1}{2} + 1 = \frac{50}{2} + 1 = 25 + 1 = 26$.

Thus, there are 26 terms in the simplified expansion.

The total number of terms in the expansion of $(x + a)^{51} – (x – a)^{51}$ after simplification is 26.

The correct option is (c) 26.

Given:

The expansion is $\left( 2 + \frac{x}{3} \right)^n$.

The coefficients of $x^7$ and $x^8$ in the expansion are equal.

To Find:

The value of $n$.

Solution:

The given expansion is of the form $(a+b)^n$, where $a=2$, $b=\frac{x}{3}$.

The general term, $T_{k+1}$, in the expansion of $(a+b)^n$ is given by $T_{k+1} = \binom{n}{k} a^{n-k} b^k$.

Substituting the values $a=2$ and $b=\frac{x}{3}$:

$T_{k+1} = \binom{n}{k} (2)^{n-k} \left(\frac{x}{3}\right)^k$

$T_{k+1} = \binom{n}{k} 2^{n-k} \frac{x^k}{3^k}$

$T_{k+1} = \binom{n}{k} 2^{n-k} 3^{-k} x^k$

The coefficient of $x^k$ in the expansion is $\binom{n}{k} 2^{n-k} 3^{-k}$.

The coefficient of $x^7$ is obtained by setting $k=7$:

Coefficient of $x^7 = \binom{n}{7} 2^{n-7} 3^{-7}$

The coefficient of $x^8$ is obtained by setting $k=8$:

Coefficient of $x^8 = \binom{n}{8} 2^{n-8} 3^{-8}$

According to the problem statement, the coefficients of $x^7$ and $x^8$ are equal:

Rewrite the binomial coefficients and powers:

$\frac{n!}{7!(n-7)!} \cdot 2^{n-7} \cdot \frac{1}{3^7} = \frac{n!}{8!(n-8)!} \cdot 2^{n-8} \cdot \frac{1}{3^8}$

Expand factorials $8! = 8 \cdot 7!$ and $(n-7)! = (n-7)(n-8)!$ (assuming $n \ge 8$):

$\frac{n!}{7!(n-7)(n-8)!} \cdot 2^{n-7} \cdot \frac{1}{3^7} = \frac{n!}{8 \cdot 7!(n-8)!} \cdot 2^{n-8} \cdot \frac{1}{3^8}$

Cancel common terms $n!$, $7!$, $(n-8)!$ from both sides:

$\frac{1}{n-7} \cdot 2^{n-7} \cdot \frac{1}{3^7} = \frac{1}{8} \cdot 2^{n-8} \cdot \frac{1}{3^8}$

Rearrange terms:

$\frac{2^{n-7}}{3^7 (n-7)} = \frac{2^{n-8}}{3^8 \cdot 8}$

Divide both sides by $2^{n-8}$ and multiply by $3^8$:

$\frac{2^{n-7}}{2^{n-8}} \cdot \frac{3^8}{3^7} = \frac{n-7}{8}$

$2^{n-7 - (n-8)} \cdot 3^{8-7} = \frac{n-7}{8}$

$2^1 \cdot 3^1 = \frac{n-7}{8}$

$6 = \frac{n-7}{8}$

Multiply both sides by 8:

$6 \times 8 = n-7$

$48 = n-7$

$n = 48 + 7$

$n = 55$

The value of $n=55$ satisfies the condition $n \ge 8$ required for $\binom{n}{8}$ to be defined in this context.

The value of $n$ is 55.

The correct option is (b) 55.

Given:

The expansion $(1 – x + x^2)^n = a_0 + a_1x + a_2x^2 + \dots + a_{2n} x^{2n}$.

To Find:

The value of $a_0 + a_2 + a_4 + \dots + a_{2n}$.

Solution:

We are given the polynomial identity:

$(1 – x + x^2)^n = a_0 + a_1x + a_2x^2 + a_3x^3 + \dots + a_{2n-1}x^{2n-1} + a_{2n} x^{2n}$

To find the sum of coefficients with even indices ($a_0 + a_2 + a_4 + \dots$), we can use properties of polynomial evaluations.

Substitute $x=1$ into the identity:

$(1 – 1 + 1^2)^n = a_0 + a_1(1) + a_2(1)^2 + a_3(1)^3 + \dots + a_{2n}(1)^{2n}$

$(1 – 1 + 1)^n = a_0 + a_1 + a_2 + a_3 + \dots + a_{2n}$

$1^n = a_0 + a_1 + a_2 + a_3 + \dots + a_{2n}$

Substitute $x=-1$ into the identity:

$(1 – (-1) + (-1)^2)^n = a_0 + a_1(-1) + a_2(-1)^2 + a_3(-1)^3 + \dots + a_{2n}(-1)^{2n}$

$(1 + 1 + 1)^n = a_0 - a_1 + a_2 - a_3 + \dots + a_{2n-1}(-1)^{2n-1} + a_{2n}(-1)^{2n}$

Since $2n$ is even, $(-1)^{2n} = 1$. Since $2n-1$ is odd, $(-1)^{2n-1} = -1$. In general, $(-1)^k$ is 1 if $k$ is even and -1 if $k$ is odd.

$3^n = a_0 - a_1 + a_2 - a_3 + \dots - a_{2n-1} + a_{2n}$

We want to find $a_0 + a_2 + a_4 + \dots + a_{2n}$, which is the sum of coefficients with even indices.

Add equation (i) and equation (ii):

$(a_0 + a_1 + a_2 + a_3 + \dots + a_{2n}) + (a_0 - a_1 + a_2 - a_3 + \dots + a_{2n}) = 1 + 3^n$

Combine like terms. The coefficients with odd indices ($a_1, a_3, \dots$) cancel out:

$(a_0 + a_0) + (a_1 - a_1) + (a_2 + a_2) + (a_3 - a_3) + \dots + (a_{2n} + a_{2n}) = 1 + 3^n$

$2a_0 + 0 + 2a_2 + 0 + \dots + 2a_{2n} = 1 + 3^n$

$2(a_0 + a_2 + a_4 + \dots + a_{2n}) = 1 + 3^n$

Divide by 2:

$a_0 + a_2 + a_4 + \dots + a_{2n} = \frac{1 + 3^n}{2}$

The value of $a_0 + a_2 + a_4 + \dots + a_{2n}$ is $\frac{3^n + 1}{2}$.

The correct option is (A) $\frac{3^n + 1}{2}$.

Given:

The expansion is $(1 + x)^{p + q}$, where $p$ and $q$ are positive integers.

To Find:

The relationship between the coefficient of $x^p$ and $x^q$ in the expansion.

Solution:

The binomial expansion of $(1+x)^n$ is given by:

$(1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k$

The coefficient of $x^k$ in the expansion of $(1+x)^n$ is $\binom{n}{k}$.

In the given expansion $(1 + x)^{p + q}$, the value of $n$ is $p+q$.

The coefficient of $x^p$ is obtained by setting $k=p$. The coefficient is $\binom{p+q}{p}$.

The coefficient of $x^q$ is obtained by setting $k=q$. The coefficient is $\binom{p+q}{q}$.

We need to compare $\binom{p+q}{p}$ and $\binom{p+q}{q}$.

Using the symmetry property of binomial coefficients, $\binom{N}{K} = \binom{N}{N-K}$, where $N$ and $K$ are non-negative integers with $0 \leq K \leq N$.

In our case, let $N = p+q$ and $K = p$. Since $p$ and $q$ are positive integers, $p+q$ is a positive integer, and $0 \leq p \leq p+q$ is true.

Applying the symmetry property:

$\binom{p+q}{p} = \binom{p+q}{(p+q) - p}$

$\binom{p+q}{p} = \binom{p+q}{q}$

This shows that the coefficient of $x^p$ is equal to the coefficient of $x^q$.

The coefficient of $x^p$ and $x^q$ in the expansion of $(1 + x)^{p + q}$ are equal.

The correct option is (A) equal.

Given:

The expression $(a + b + c)^n$, where $n \in \mathbb{N}$.

To Find:

The number of terms in the expansion.

Solution:

The expansion of $(a_1 + a_2 + \dots + a_m)^n$ is given by the multinomial theorem:

$\sum_{n_1+n_2+\dots+n_m=n} \frac{n!}{n_1! n_2! \dots n_m!} a_1^{n_1} a_2^{n_2} \dots a_m^{n_m}$

Each term in the expansion corresponds to a unique combination of non-negative integers $(n_1, n_2, \dots, n_m)$ such that $n_1 + n_2 + \dots + n_m = n$.

In this case, we have a trinomial $(a+b+c)^n$, so $m=3$. The terms are of the form $C \cdot a^{n_1} b^{n_2} c^{n_3}$, where $C$ is the coefficient and $n_1, n_2, n_3$ are non-negative integers such that $n_1 + n_2 + n_3 = n$.

The number of distinct terms in the expansion is equal to the number of distinct non-negative integer solutions to the equation $n_1 + n_2 + n_3 = n$.

This is a classic combinatorial problem that can be solved using stars and bars. We have $n$ "stars" to distribute among 3 "bins" (corresponding to the powers $n_1, n_2, n_3$). We need $m-1 = 3-1 = 2$ "bars" to separate the bins.

The number of ways to arrange $n$ stars and 2 bars is the number of ways to choose the positions of the 2 bars (or the $n$ stars) in a sequence of $n+2$ positions.

The number of terms is $\binom{n + m - 1}{m - 1}$ or $\binom{n + m - 1}{n}$.

Using $m=3$, the number of terms is $\binom{n + 3 - 1}{3 - 1} = \binom{n+2}{2}$.

Calculate the binomial coefficient:

$\binom{n+2}{2} = \frac{(n+2)!}{2!(n+2-2)!} = \frac{(n+2)!}{2!n!} = \frac{(n+2)(n+1)\cancel{n!}}{2 \times 1 \times \cancel{n!}} = \frac{(n+2)(n+1)}{2}$

The number of terms in the expansion of $(a + b + c)^n$ is $\frac{(n+1)(n+2)}{2}$.

The correct option is (A) $\frac{(n+1) (n+2)}{2}$.

Given:

The expansion is $\left( x^2 + \frac{2}{x} \right)^{15}$.

To Find:

The ratio of the coefficient of $x^{15}$ to the term independent of $x$.

Solution:

The given expansion is of the form $(a+b)^n$, where $a=x^2$, $b=\frac{2}{x}$, and $n=15$.

The general term, $T_{k+1}$, in the expansion is given by $T_{k+1} = \binom{n}{k} a^{n-k} b^k$.

Substituting the values $a=x^2$, $b=\frac{2}{x}$, and $n=15$:

$T_{k+1} = \binom{15}{k} (x^2)^{15-k} \left(\frac{2}{x}\right)^k$

Simplify the terms involving $x$:

$(x^2)^{15-k} = x^{2(15-k)} = x^{30-2k}$

$\left(\frac{2}{x}\right)^k = \frac{2^k}{x^k} = 2^k x^{-k}$

The variable part of the general term is $x^{30-2k} \cdot x^{-k} = x^{30-2k-k} = x^{30-3k}$.

The general term is $T_{k+1} = \binom{15}{k} 2^k x^{30-3k}$.

Coefficient of $x^{15}$:

For the coefficient of $x^{15}$, we set the power of $x$ equal to 15:

$30 - 3k = 15$

$30 - 15 = 3k$

$15 = 3k$

$k = 5$

The coefficient of $x^{15}$ is the constant part of the general term when $k=5$:

Coefficient of $x^{15} = \binom{15}{5} 2^5$

$\binom{15}{5} = \frac{15!}{5!10!} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} = \frac{\cancel{15}^{1} \times \cancel{14}^{7} \times 13 \times \cancel{12}^{1} \times 11}{\cancel{5}_{1} \times \cancel{4}_{1} \times \cancel{3}_{1} \times \cancel{2}_{1} \times 1} = 1 \times 7 \times 13 \times 1 \times 11 = 3003$

$2^5 = 32$

Coefficient of $x^{15} = 3003 \times 32 = 96096$.

Term independent of $x$:

The term independent of $x$ occurs when the power of $x$ is 0:

$30 - 3k = 0$

$30 = 3k$

$k = 10$

The term independent of $x$ is the value of the general term when $k=10$ (without the $x^0$):

Term independent of $x = \binom{15}{10} 2^{10}$

$\binom{15}{10} = \binom{15}{15-10} = \binom{15}{5} = 3003$ (as calculated before)

$2^{10} = 1024$

Term independent of $x = 3003 \times 1024 = 3075072$.

Ratio:

We need the ratio of the coefficient of $x^{15}$ to the term independent of $x$.

Ratio $= \frac{\text{Coefficient of } x^{15}}{\text{Term independent of } x} = \frac{\binom{15}{5} 2^5}{\binom{15}{10} 2^{10}}$

Since $\binom{15}{5} = \binom{15}{10}$, these terms cancel out:

Ratio $= \frac{2^5}{2^{10}} = 2^{5-10} = 2^{-5} = \frac{1}{2^5} = \frac{1}{32}$

The ratio is $1:32$.

The ratio of the coefficient of $x^{15}$ to the term independent of $x$ is $1:32$.

The correct option is (B) 1:32.

Given:

$z = \left( \frac{\sqrt{3}}{2} + \frac{i}{2} \right)^5 + \left( \frac{\sqrt{3}}{2} - \frac{i}{2} \right)^5$.

To Find:

The properties of the complex number $z$, specifically its real and imaginary parts.

Solution:

Let $a = \frac{\sqrt{3}}{2}$ and $b = \frac{1}{2}$. The expression for $z$ is in the form $(a+ib)^5 + (a-ib)^5$.

We can use the binomial theorem to expand each term. Let $n=5$.

$(a+ib)^n = \binom{n}{0}a^n(ib)^0 + \binom{n}{1}a^{n-1}(ib)^1 + \binom{n}{2}a^{n-2}(ib)^2 + \binom{n}{3}a^{n-3}(ib)^3 + \dots + \binom{n}{n}a^0(ib)^n$

$(a-ib)^n = \binom{n}{0}a^n(-ib)^0 + \binom{n}{1}a^{n-1}(-ib)^1 + \binom{n}{2}a^{n-2}(-ib)^2 + \binom{n}{3}a^{n-3}(-ib)^3 + \dots + \binom{n}{n}a^0(-ib)^n$

Using the property $(-ib)^k = (-1)^k (ib)^k$.

Summing the two expansions:

$(a+ib)^n + (a-ib)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} (ib)^k + \sum_{k=0}^n \binom{n}{k} a^{n-k} (-1)^k (ib)^k$

$= \sum_{k=0}^n \binom{n}{k} a^{n-k} (ib)^k (1 + (-1)^k)$

The term $(1 + (-1)^k)$ is 2 if $k$ is even, and 0 if $k$ is odd.

So, only the terms where $k$ is even will contribute to the sum.

For $n=5$, the possible values of $k$ are $0, 1, 2, 3, 4, 5$. The even values of $k$ are $0, 2, 4$.

$z = (a+ib)^5 + (a-ib)^5 = 2 \left[ \binom{5}{0}a^5(ib)^0 + \binom{5}{2}a^3(ib)^2 + \binom{5}{4}a^1(ib)^4 \right]$

Substitute $a=\frac{\sqrt{3}}{2}$ and $b=\frac{1}{2}$, and powers of $i$: $i^0=1, i^2=-1, i^4=1$.

$z = 2 \left[ \binom{5}{0}\left(\frac{\sqrt{3}}{2}\right)^5(1) + \binom{5}{2}\left(\frac{\sqrt{3}}{2}\right)^3(i^2 (\frac{1}{2})^2) + \binom{5}{4}\left(\frac{\sqrt{3}}{2}\right)^1(i^4 (\frac{1}{2})^4) \right]$

$z = 2 \left[ \binom{5}{0}\left(\frac{\sqrt{3}}{2}\right)^5 + \binom{5}{2}\left(\frac{\sqrt{3}}{2}\right)^3(-1 \cdot \frac{1}{4}) + \binom{5}{4}\left(\frac{\sqrt{3}}{2}\right)\left(1 \cdot \frac{1}{16}\right) \right]$

$z = 2 \left[ \binom{5}{0}\left(\frac{\sqrt{3}}{2}\right)^5 - \frac{1}{4}\binom{5}{2}\left(\frac{\sqrt{3}}{2}\right)^3 + \frac{1}{16}\binom{5}{4}\left(\frac{\sqrt{3}}{2}\right) \right]$

Calculate the binomial coefficients:

$\binom{5}{0} = 1$

$\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10$

$\binom{5}{4} = \binom{5}{1} = 5$

Calculate the powers of $\frac{\sqrt{3}}{2}$:

$\left(\frac{\sqrt{3}}{2}\right)^1 = \frac{\sqrt{3}}{2}$

$\left(\frac{\sqrt{3}}{2}\right)^3 = \frac{(\sqrt{3})^3}{2^3} = \frac{3\sqrt{3}}{8}$

$\left(\frac{\sqrt{3}}{2}\right)^5 = \frac{(\sqrt{3})^5}{2^5} = \frac{9\sqrt{3}}{32}$

Substitute these values into the expression for $z$:

$z = 2 \left[ 1 \cdot \frac{9\sqrt{3}}{32} - \frac{1}{4} \cdot 10 \cdot \frac{3\sqrt{3}}{8} + \frac{1}{16} \cdot 5 \cdot \frac{\sqrt{3}}{2} \right]$

$z = 2 \left[ \frac{9\sqrt{3}}{32} - \frac{10 \cdot 3\sqrt{3}}{32} + \frac{5\sqrt{3}}{32} \right]$

$z = 2 \left[ \frac{9\sqrt{3} - 30\sqrt{3} + 5\sqrt{3}}{32} \right]$

$z = 2 \left[ \frac{(9 - 30 + 5)\sqrt{3}}{32} \right]$

$z = 2 \left[ \frac{-16\sqrt{3}}{32} \right]$

$z = 2 \left[ -\frac{\cancel{16}\sqrt{3}}{\cancel{32}_{2}} \right]$

$z = 2 \left[ -\frac{\sqrt{3}}{2} \right]$

$z = -\sqrt{3}$

The complex number $z = -\sqrt{3}$.

The real part of $z$ is $\text{Re}(z) = -\sqrt{3}$.

The imaginary part of $z$ is $\text{Im}(z) = 0$.

Since $\sqrt{3} \approx 1.732$, $\text{Re}(z) = -\sqrt{3} < 0$.

$\text{Im}(z) = 0$.

Based on the options:

(A) $\text{Re}(z) = 0$ (False)

(B) $\text{Im}(z) = 0$ (True)

(C) $\text{Re}(z) > 0$, $\text{Im}(z) > 0$ (False)

(D) $\text{Re}(z) > 0$, $\text{Im}(z) < 0$ (False)

The imaginary part of $z$ is 0.

The correct option is (B) I_m (z) = 0.

Alternate Solution using Polar Form:

The complex number $\frac{\sqrt{3}}{2} + \frac{i}{2}$ can be written in polar form $r(\cos \theta + i \sin \theta)$.

$r = \left| \frac{\sqrt{3}}{2} + \frac{i}{2} \right| = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{\frac{4}{4}} = \sqrt{1} = 1$

$\cos \theta = \frac{\sqrt{3}}{2}$ and $\sin \theta = \frac{1}{2}$. This means $\theta = \frac{\pi}{6}$ (or $30^\circ$).

So, $\frac{\sqrt{3}}{2} + \frac{i}{2} = 1 \cdot (\cos(\frac{\pi}{6}) + i \sin(\frac{\pi}{6})) = e^{i \pi/6}$.

The complex conjugate is $\frac{\sqrt{3}}{2} - \frac{i}{2} = \cos(\frac{\pi}{6}) - i \sin(\frac{\pi}{6}) = \cos(-\frac{\pi}{6}) + i \sin(-\frac{\pi}{6}) = e^{-i \pi/6}$.

Using De Moivre's Theorem, $(r(\cos \theta + i \sin \theta))^n = r^n(\cos(n\theta) + i \sin(n\theta))$.

$\left( \frac{\sqrt{3}}{2} + \frac{i}{2} \right)^5 = (e^{i \pi/6})^5 = e^{i 5\pi/6} = \cos(\frac{5\pi}{6}) + i \sin(\frac{5\pi}{6})$

$\left( \frac{\sqrt{3}}{2} - \frac{i}{2} \right)^5 = (e^{-i \pi/6})^5 = e^{-i 5\pi/6} = \cos(-\frac{5\pi}{6}) + i \sin(-\frac{5\pi}{6})$

Using $\cos(-x) = \cos(x)$ and $\sin(-x) = -\sin(x)$:

$\left( \frac{\sqrt{3}}{2} - \frac{i}{2} \right)^5 = \cos(\frac{5\pi}{6}) - i \sin(\frac{5\pi}{6})$

Now, sum the two terms to find $z$:

$z = \left( \cos(\frac{5\pi}{6}) + i \sin(\frac{5\pi}{6}) \right) + \left( \cos(\frac{5\pi}{6}) - i \sin(\frac{5\pi}{6}) \right)$

$z = 2 \cos(\frac{5\pi}{6})$

The value of $\cos(\frac{5\pi}{6}) = \cos(\pi - \frac{\pi}{6}) = -\cos(\frac{\pi}{6}) = -\frac{\sqrt{3}}{2}$.

$z = 2 \left(-\frac{\sqrt{3}}{2}\right) = -\sqrt{3}$

So, $z = -\sqrt{3} + 0i$.

$\text{Re}(z) = -\sqrt{3}$ and $\text{Im}(z) = 0$.

This confirms that the imaginary part of $z$ is 0.

The correct option is (B) I_m (z) = 0.

Given:

The binomial expansion is $\left( \frac{3x^2}{2} − \frac{1}{3x} \right)^{15}$, with $x \neq 0$.

To Find:

The term independent of $x$ in the expansion.

Solution:

The given expansion is of the form $(a+b)^n$, where $a = \frac{3x^2}{2}$, $b = -\frac{1}{3x}$, and $n = 15$.

The general term, $T_{k+1}$, in the expansion of $(a+b)^n$ is given by the formula:

$T_{k+1} = \binom{n}{k} a^{n-k} b^k$

Substitute the values of $n$, $a$, and $b$ for the given expansion:

$T_{k+1} = \binom{15}{k} \left(\frac{3x^2}{2}\right)^{15-k} \left(-\frac{1}{3x}\right)^k$

Now, let's separate the constant terms and the terms involving $x$.

Constant part: $\binom{15}{k} \left(\frac{3}{2}\right)^{15-k} \left(-\frac{1}{3}\right)^k$

Variable part: $(x^2)^{15-k} \cdot \left(\frac{1}{x}\right)^k = x^{2(15-k)} \cdot x^{-k} = x^{30-2k} \cdot x^{-k} = x^{30-3k}$

The general term is $T_{k+1} = \binom{15}{k} \left(\frac{3}{2}\right)^{15-k} \left(-\frac{1}{3}\right)^k x^{30-3k}$.

For the term independent of $x$, the power of $x$ must be zero. So, we set the exponent of $x$ equal to 0:

$30 - 3k = 0$

Solve for $k$:

$30 = 3k$

$k = 10$

Since $k=10$ is a non-negative integer and $0 \leq 10 \leq 15$, this value of $k$ is valid.

The term independent of $x$ is the constant part of $T_{k+1}$ when $k=10$.

Term independent of $x = \binom{15}{10} \left(\frac{3}{2}\right)^{15-10} \left(-\frac{1}{3}\right)^{10}$

Term independent of $x = \binom{15}{10} \left(\frac{3}{2}\right)^5 \left(-\frac{1}{3}\right)^{10}$

Calculate the binomial coefficient $\binom{15}{10}$:

$\binom{15}{10} = \binom{15}{15-10} = \binom{15}{5} = \frac{15!}{5!10!} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} = 3003$

Calculate the powers of the fractions:

$\left(\frac{3}{2}\right)^5 = \frac{3^5}{2^5} = \frac{243}{32}$

$\left(-\frac{1}{3}\right)^{10} = (-1)^{10} \frac{1^{10}}{3^{10}} = 1 \cdot \frac{1}{3^{10}} = \frac{1}{59049}$

Substitute these values back into the expression for the term independent of $x$:

Term independent of $x = 3003 \cdot \frac{243}{32} \cdot \frac{1}{59049}$

Simplify the numerical part. Note that $59049 = 3^{10}$ and $243 = 3^5$.

Term independent of $x = 3003 \cdot \frac{3^5}{32} \cdot \frac{1}{3^{10}}$

Term independent of $x = 3003 \cdot \frac{1}{32} \cdot \frac{3^5}{3^{10}}$

Term independent of $x = 3003 \cdot \frac{1}{32} \cdot 3^{5-10}$

Term independent of $x = 3003 \cdot \frac{1}{32} \cdot 3^{-5}$

Term independent of $x = 3003 \cdot \frac{1}{32} \cdot \frac{1}{3^5}$

Term independent of $x = 3003 \cdot \frac{1}{32} \cdot \frac{1}{243}$

Term independent of $x = \frac{3003}{32 \times 243}$

Term independent of $x = \frac{3003}{7776}$

Let's check if there are common factors. The sum of digits of 3003 is $3+0+0+3=6$, which is divisible by 3. So 3003 is divisible by 3. $3003 / 3 = 1001$.

The sum of digits of 7776 is $7+7+7+6=27$, which is divisible by 9 (and 3). So 7776 is divisible by 3. $7776 / 3 = 2592$.

$\frac{3003}{7776} = \frac{1001}{2592}$

Let's express $\binom{15}{10} \frac{3^5}{2^5} \frac{1}{3^{10}}$ as $\frac{15!}{10!5!} \frac{3^5}{2^5 3^{10}} = \frac{15!}{10!5!} \frac{1}{2^5 3^5}$.

$\frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} \times \frac{1}{32 \times 243} = 3003 \times \frac{1}{32 \times 243} = \frac{3003}{7776}$.

Is 1001 divisible by any factors of 2592 other than 1? $2592 = 2^5 \times 3^4$. 1001 is not divisible by 2 or 3. So the fraction is $\frac{1001}{2592}$.

The term independent of $x$ is $\frac{1001}{2592}$.

The final answer is $\boxed{\frac{1001}{2592}}$.

Given:

The expansion is $\left( \sqrt{x} - \frac{k}{x^2} \right)^{10}$.

The term independent of $x$ is 405.

To Find:

The value of $k$.

Solution:

The given expansion is of the form $(a+b)^n$, where $a = \sqrt{x} = x^{1/2}$, $b = -\frac{k}{x^2} = -k x^{-2}$, and $n = 10$.

The general term, $T_{r+1}$, in the expansion is given by the formula:

$T_{r+1} = \binom{n}{r} a^{n-r} b^r$

Substitute the values of $n$, $a$, and $b$ for the given expansion:

$T_{r+1} = \binom{10}{r} (x^{1/2})^{10-r} (-k x^{-2})^r$

Simplify the terms involving $x$ and the constant terms separately.

Constant part: $\binom{10}{r} (-k)^r = \binom{10}{r} (-1)^r k^r$

Variable part: $(x^{1/2})^{10-r} \cdot (x^{-2})^r = x^{(1/2)(10-r)} \cdot x^{-2r} = x^{\frac{10-r}{2}} \cdot x^{-2r} = x^{\frac{10-r}{2} - 2r} = x^{\frac{10-r-4r}{2}} = x^{\frac{10-5r}{2}}$

The general term is $T_{r+1} = \binom{10}{r} (-1)^r k^r x^{\frac{10-5r}{2}}$.

For the term independent of $x$, the power of $x$ must be zero. So, we set the exponent of $x$ equal to 0:

$\frac{10-5r}{2} = 0$

$10 - 5r = 0$

$10 = 5r$

$r = 2$

Since $r=2$ is a non-negative integer and $0 \leq 2 \leq 10$, this value of $r$ is valid.

The term independent of $x$ is the constant part of $T_{r+1}$ when $r=2$.

Term independent of $x = \binom{10}{2} (-1)^2 k^2$

We are given that this term is equal to 405.

$\binom{10}{2} (-1)^2 k^2 = 405$

Calculate the binomial coefficient $\binom{10}{2}$:

$\binom{10}{2} = \frac{10!}{2!8!} = \frac{10 \times 9}{2 \times 1} = 45$

$(-1)^2 = 1$

Substitute these values into the equation:

$45 \cdot 1 \cdot k^2 = 405$

$45k^2 = 405$

Divide both sides by 45:

$k^2 = \frac{405}{45}$

Divide 405 by 45. $405 = 9 \times 45$.

$k^2 = \frac{9 \times \cancel{45}}{\cancel{45}} = 9$

Take the square root of both sides:

$k = \pm \sqrt{9}$

$k = \pm 3$

The possible values for $k$ are 3 and -3.

The final answer is $\boxed{\pm 3}$.

Given:

The expression $(1 – 3x + 7x^2) (1 – x)^{16}$.

To Find:

The coefficient of $x$ in the expansion of the given expression.

Solution:

Let the given expression be $E$.

$E = (1 – 3x + 7x^2) (1 – x)^{16}$

We need to find the terms that produce $x^1$ when the two factors are multiplied.

The second factor is $(1-x)^{16}$. The binomial expansion of $(1+y)^n = \sum_{k=0}^n \binom{n}{k} y^k$.

Here $y=-x$ and $n=16$.

$(1-x)^{16} = \binom{16}{0}(1)^{16}(-x)^0 + \binom{16}{1}(1)^{15}(-x)^1 + \binom{16}{2}(1)^{14}(-x)^2 + \dots$

$(1-x)^{16} = \binom{16}{0} - \binom{16}{1}x + \binom{16}{2}x^2 - \binom{16}{3}x^3 + \dots$

$(1-x)^{16} = 1 - 16x + \binom{16}{2}x^2 - \dots$

The first factor is $(1 – 3x + 7x^2)$.

We multiply the two factors:

$E = (1 – 3x + 7x^2) (1 - 16x + \binom{16}{2}x^2 - \dots)$

To get a term with $x^1$, we multiply terms from the first factor with terms from the second factor such that the powers of $x$ add up to 1.

The possible combinations are:

1. Constant term from the first factor multiplied by the $x^1$ term from the second factor.

$(1) \cdot (-\binom{16}{1}x) = -16x$

2. $x^1$ term from the first factor multiplied by the constant term from the second factor.

$(-3x) \cdot (\binom{16}{0}) = -3x \cdot 1 = -3x$

The terms with $x^2$ or higher powers from the first factor, when multiplied by any term from the second factor up to the $x^0$ term, will result in a power of $x$ greater than or equal to 2. For example, $(7x^2) \cdot (1) = 7x^2$.

Therefore, the terms containing $x^1$ in the expansion are $-16x$ and $-3x$.

The coefficient of $x$ is the sum of the coefficients of these terms.

Coefficient of $x = -16 + (-3) = -16 - 3 = -19$.

The coefficient of $x$ in the expansion of $(1 – 3x + 7x^2) (1 – x)^{16}$ is $-19$.

The final answer is $\boxed{-19}$.

Given:

The binomial expansion is $\left( 3x - \frac{2}{x^2} \right)^{15}$.

To Find:

The term independent of $x$ in the expansion.

Solution:

The given expansion is of the form $(a+b)^n$, where $a = 3x$, $b = -\frac{2}{x^2}$, and $n = 15$.

The general term, $T_{k+1}$, in the expansion of $(a+b)^n$ is given by the formula:

$T_{k+1} = \binom{n}{k} a^{n-k} b^k$

Substitute the values of $n$, $a$, and $b$ for the given expansion:

$T_{k+1} = \binom{15}{k} (3x)^{15-k} \left(-\frac{2}{x^2}\right)^k$

Now, let's separate the constant terms and the terms involving $x$.

Constant part: $\binom{15}{k} 3^{15-k} (-2)^k$

Variable part: $(x)^{15-k} \cdot \left(\frac{1}{x^2}\right)^k = x^{15-k} \cdot (x^{-2})^k = x^{15-k} \cdot x^{-2k} = x^{15-k-2k} = x^{15-3k}$

The general term is $T_{k+1} = \binom{15}{k} 3^{15-k} (-2)^k x^{15-3k}$.

For the term independent of $x$, the power of $x$ must be zero. So, we set the exponent of $x$ equal to 0:

$15 - 3k = 0$

Solve for $k$:

$15 = 3k$

$k = 5$

Since $k=5$ is a non-negative integer and $0 \leq 5 \leq 15$, this value of $k$ is valid.

The term independent of $x$ is the constant part of $T_{k+1}$ when $k=5$.

Term independent of $x = \binom{15}{5} 3^{15-5} (-2)^5$

Term independent of $x = \binom{15}{5} 3^{10} (-2)^5$

Calculate the binomial coefficient $\binom{15}{5}$:

$\binom{15}{5} = \frac{15!}{5!10!} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} = 3003$

Calculate the powers:

$3^{10} = 59049$

$(-2)^5 = -32$

Substitute these values back into the expression for the term independent of $x$:

Term independent of $x = 3003 \cdot 59049 \cdot (-32)$

Term independent of $x = 3003 \times 59049 \times (-32) = 177347947 \times (-32) = -5675134304$

The term independent of $x$ in the expansion is $-5675134304$.

The final answer is $\boxed{-5675134304}$.

Given: Expansions (i) $\left( \frac{x}{a} − \frac{a}{x} \right)^{10}$ and (ii) $\left( 3x - \frac{x^3}{6} \right)^9$.

To Find: Middle term(s).

Solution:

The general term $T_{k+1}$ of $(A+B)^n$ is $\binom{n}{k} A^{n-k} B^k$.

(i) $\left( \frac{x}{a} − \frac{a}{x} \right)^{10}$: Here $n=10$ (even). Total terms = 11. Middle term position = $\frac{10}{2} + 1 = 6^{\text{th}}$. So $k=5$.

$T_6 = \binom{10}{5} \left(\frac{x}{a}\right)^{10-5} \left(-\frac{a}{x}\right)^5 = \binom{10}{5} \left(\frac{x}{a}\right)^5 \left(-\frac{a}{x}\right)^5$

$T_6 = \binom{10}{5} \frac{x^5}{a^5} (-1)^5 \frac{a^5}{x^5} = \binom{10}{5} (-1)$

$\binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$.

$T_6 = 252 \times (-1) = -252$.

(ii) $\left( 3x - \frac{x^3}{6} \right)^9$: Here $n=9$ (odd). Total terms = 10. Middle term positions = $\frac{9+1}{2} = 5^{\text{th}}$ and $6^{\text{th}}$.

For $T_5$, $k=4$. $A=3x$, $B=-\frac{x^3}{6}$.

$T_5 = \binom{9}{4} (3x)^{9-4} \left(-\frac{x^3}{6}\right)^4 = \binom{9}{4} (3x)^5 \left(-\frac{x^3}{6}\right)^4$

$T_5 = \binom{9}{4} 3^5 x^5 (-1)^4 \frac{(x^3)^4}{6^4} = \binom{9}{4} 3^5 x^5 \frac{x^{12}}{6^4} = \binom{9}{4} \frac{3^5}{6^4} x^{17}$

$\binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$.

$\frac{3^5}{6^4} = \frac{3^5}{(2 \times 3)^4} = \frac{3^5}{2^4 \times 3^4} = \frac{3}{2^4} = \frac{3}{16}$.

$T_5 = 126 \cdot \frac{3}{16} x^{17} = \frac{\cancel{126}^{63} \times 3}{\cancel{16}_{8}} x^{17} = \frac{189}{8} x^{17}$.

For $T_6$, $k=5$.

$T_6 = \binom{9}{5} (3x)^{9-5} \left(-\frac{x^3}{6}\right)^5 = \binom{9}{5} (3x)^4 \left(-\frac{x^3}{6}\right)^5$

$T_6 = \binom{9}{5} 3^4 x^4 (-1)^5 \frac{(x^3)^5}{6^5} = \binom{9}{5} 3^4 x^4 (-1) \frac{x^{15}}{6^5} = -\binom{9}{5} \frac{3^4}{6^5} x^{19}$

$\binom{9}{5} = \binom{9}{4} = 126$.

$\frac{3^4}{6^5} = \frac{3^4}{(2 \times 3)^5} = \frac{3^4}{2^5 \times 3^5} = \frac{1}{2^5 \times 3} = \frac{1}{32 \times 3} = \frac{1}{96}$.

$T_6 = -126 \cdot \frac{1}{96} x^{19} = -\frac{\cancel{126}^{21}}{\cancel{96}_{16}} x^{19} = -\frac{21}{16} x^{19}$.

The middle term for (i) is -252.

The middle terms for (ii) are $\frac{189}{8} x^{17}$ and $-\frac{21}{16} x^{19}$.

The final answer is $\boxed{\text{(i) -252, (ii) } \frac{189}{8} x^{17}, -\frac{21}{16} x^{19}}$.

Given:

The expansion is $(x – x^2)^{10}$.

To Find:

The coefficient of $x^{15}$ in the expansion.

Solution:

We can factor out $x$ from the base of the binomial:

$(x – x^2)^{10} = (x(1 – x))^{10} = x^{10} (1 – x)^{10}$

Now, expand $(1-x)^{10}$ using the binomial theorem. The general term in the expansion of $(1+y)^n$ is $\binom{n}{k} y^k$. Here $y=-x$ and $n=10$.

$(1-x)^{10} = \sum_{k=0}^{10} \binom{10}{k} (-x)^k = \sum_{k=0}^{10} \binom{10}{k} (-1)^k x^k$

Substitute this back into the expression:

$(x – x^2)^{10} = x^{10} \sum_{k=0}^{10} \binom{10}{k} (-1)^k x^k$

Multiply $x^{10}$ with each term in the sum:

$(x – x^2)^{10} = \sum_{k=0}^{10} \binom{10}{k} (-1)^k x^{10} x^k$

Using the property $x^p x^q = x^{p+q}$:

$(x – x^2)^{10} = \sum_{k=0}^{10} \binom{10}{k} (-1)^k x^{10+k}$

We want to find the coefficient of $x^{15}$. This means the power of $x$ must be equal to 15.

So, we set the exponent of $x$ equal to 15:

$10 + k = 15$

Solve for $k$:

$k = 15 - 10$

$k = 5$

The index $k$ in the summation ranges from 0 to 10. Since $k=5$ is within this range ($0 \leq 5 \leq 10$), there is a term with $x^{15}$.

The coefficient of $x^{15}$ is the coefficient of the term where $k = 5$, which is $\binom{10}{k} (-1)^k$ for $k=5$.

Coefficient of $x^{15} = \binom{10}{5} (-1)^5$

Calculate the binomial coefficient $\binom{10}{5}$:

$\binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$

Calculate $(-1)^5$:

$(-1)^5 = -1$

Multiply the results:

Coefficient of $x^{15} = 252 \times (-1) = -252$

The coefficient of $x^{15}$ in the expansion of $(x – x^2)^{10}$ is $-252$.

The final answer is $\boxed{-252}$.

Given:

The expansion is $\left( x^4 - \frac{1}{x^3} \right)^{15}$.

To Find:

The coefficient of $\frac{1}{x^{17}}$ (which is $x^{-17}$) in the expansion.

Solution:

The given expansion is of the form $(a+b)^n$, where $a = x^4$, $b = -\frac{1}{x^3} = -x^{-3}$, and $n = 15$.

The general term, $T_{k+1}$, in the expansion of $(a+b)^n$ is given by the formula:

$T_{k+1} = \binom{n}{k} a^{n-k} b^k$

Substitute the values of $n$, $a$, and $b$ for the given expansion:

$T_{k+1} = \binom{15}{k} (x^4)^{15-k} (-x^{-3})^k$

Simplify the terms involving $x$ and the constant terms separately.

Constant part: $\binom{15}{k} (-1)^k$

Variable part: $(x^4)^{15-k} \cdot (x^{-3})^k = x^{4(15-k)} \cdot x^{-3k} = x^{60-4k} \cdot x^{-3k} = x^{60-4k-3k} = x^{60-7k}$

The general term is $T_{k+1} = \binom{15}{k} (-1)^k x^{60-7k}$.

For the coefficient of $x^{-17}$, we set the power of $x$ equal to $-17$:

$60 - 7k = -17$

Solve for $k$:

$60 + 17 = 7k$

$77 = 7k$

$k = \frac{77}{7}$

$k = 11$

The index $k$ in the general term $T_{k+1}$ must be a non-negative integer, where $0 \leq k \leq n$. In this case, $n=15$, so $k$ must be an integer such that $0 \leq k \leq 15$.

Since $k=11$ is within this range ($0 \leq 11 \leq 15$), there is a term with $x^{-17}$.

The coefficient of $x^{-17}$ is the constant part of the general term when $k=11$.

Coefficient of $x^{-17} = \binom{15}{11} (-1)^{11}$

Calculate the binomial coefficient $\binom{15}{11}$:

$\binom{15}{11} = \binom{15}{15-11} = \binom{15}{4}$

$\binom{15}{4} = \frac{15!}{4!11!} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = \frac{\cancel{15}^{5} \times \cancel{14}^{7} \times 13 \times \cancel{12}^{1}}{\cancel{24}_{1}} = 5 \times 7 \times 13 \times 1 = 455$

Calculate $(-1)^{11}$:

$(-1)^{11} = -1$

Multiply the results:

Coefficient of $x^{-17} = 455 \times (-1) = -455$

The coefficient of $\frac{1}{x^{17}}$ in the expansion is $-455$.

The final answer is $\boxed{-455}$.

Given:

The expansion is $\left( y^{\frac{1}{2}} + x^{\frac{1}{3}} \right)^n$.

The binomial coefficient of the third term from the end is 45.

To Find:

The sixth term of the expansion.

Solution:

The given expansion is of the form $(a+b)^n$, where $a = y^{1/2}$ and $b = x^{1/3}$.

The total number of terms in the expansion is $n+1$.

The third term from the end is the $( (n+1) - 3 + 1 )^{\text{th}}$ term from the beginning.

Third term from the end is the $(n - 1)^{\text{th}}$ term from the beginning.

The coefficient of the $(r+1)^{\text{th}}$ term from the beginning in $(a+b)^n$ is $\binom{n}{r}$.

The coefficient of the $(n-1)^{\text{th}}$ term from the beginning corresponds to $r+1 = n-1$, so $r = n-2$.

The binomial coefficient of the third term from the end is $\binom{n}{n-2}$.

Using the property $\binom{n}{k} = \binom{n}{n-k}$, we have $\binom{n}{n-2} = \binom{n}{n-(n-2)} = \binom{n}{2}$.

We are given that this coefficient is 45.

$\binom{n}{2} = 45$

$\frac{n(n-1)}{2!} = 45$

$\frac{n(n-1)}{2} = 45$

$n(n-1) = 45 \times 2$

$n(n-1) = 90$

We need to find two consecutive integers whose product is 90. By inspection, $10 \times 9 = 90$.

So, $n=10$ (and $n-1=9$). Since $n$ is the power of the binomial, it must be a positive integer. $n=10$ is a valid value.

Now that we know $n=10$, the expansion is $\left( y^{\frac{1}{2}} + x^{\frac{1}{3}} \right)^{10}$.

We need to find the sixth term of this expansion.

The sixth term is $T_6$. For $T_6$, we set $k+1=6$ in the general term formula $T_{k+1} = \binom{n}{k} a^{n-k} b^k$. So $k=5$.

Here $n=10$, $a = y^{1/2}$, $b = x^{1/3}$.

$T_6 = \binom{10}{5} (y^{1/2})^{10-5} (x^{1/3})^5$

$T_6 = \binom{10}{5} (y^{1/2})^5 (x^{1/3})^5$

$T_6 = \binom{10}{5} y^{(1/2) \times 5} x^{(1/3) \times 5}$

$T_6 = \binom{10}{5} y^{5/2} x^{5/3}$

Calculate the binomial coefficient $\binom{10}{5}$:

$\binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$

Substitute the value back into the expression for $T_6$:

$T_6 = 252 y^{5/2} x^{5/3}$

The sixth term of the expansion is $252 y^{5/2} x^{5/3}$.

The final answer is $\boxed{252 y^{5/2} x^{5/3}}$.

Given:

The expansion is $(1 + x)^{18}$.

The coefficients of the $(2r + 4)^{\text{th}}$ and $(r – 2)^{\text{th}}$ terms are equal.

To Find:

The value of $r$.

Solution:

The expansion is $(1+x)^{18}$, so $n=18$.

The coefficient of the $m^{\text{th}}$ term in the expansion of $(1+x)^n$ is $\binom{n}{m-1}$.

The coefficient of the $(2r+4)^{\text{th}}$ term is $\binom{18}{(2r+4)-1} = \binom{18}{2r+3}$.

The coefficient of the $(r-2)^{\text{th}}$ term is $\binom{18}{(r-2)-1} = \binom{18}{r-3}$.

We are given that these coefficients are equal:

For the binomial coefficients $\binom{n}{a}$ and $\binom{n}{b}$ to be equal, there are two possibilities:

1. $a = b$

2. $a + b = n$

Case 1: $2r + 3 = r - 3$

$2r - r = -3 - 3$

$r = -6$

The term number must be a positive integer. The $(r-2)^{\text{th}}$ term means $r-2 \geq 1$, so $r \geq 3$. Also, the index in the binomial coefficient must be non-negative, $r-3 \geq 0$, so $r \geq 3$. The $(2r+4)^{\text{th}}$ term means $2r+4 \geq 1$, and $2r+3 \geq 0$, so $2r \geq -3$, $r \geq -3/2$. The value $r=-6$ does not satisfy $r \geq 3$. So this case is not possible.

Case 2: $(2r + 3) + (r - 3) = 18$

$2r + 3 + r - 3 = 18$

$3r = 18$

$r = \frac{18}{3}$

$r = 6$

Check if this value of $r$ is valid for the term numbers and binomial coefficient indices:

$(2r+4)^{\text{th}}$ term: $2(6)+4 = 12+4 = 16^{\text{th}}$ term. The index is $2r+3 = 2(6)+3 = 12+3 = 15$. $\binom{18}{15}$ is valid ($0 \leq 15 \leq 18$).

$(r-2)^{\text{th}}$ term: $6-2 = 4^{\text{th}}$ term. The index is $r-3 = 6-3 = 3$. $\binom{18}{3}$ is valid ($0 \leq 3 \leq 18$).

The coefficients are $\binom{18}{15}$ and $\binom{18}{3}$. We know $\binom{18}{15} = \binom{18}{18-15} = \binom{18}{3}$, so they are indeed equal.

The value $r=6$ is a valid solution.

The value of $r$ is 6.

The final answer is $\boxed{6}$.

Given:

The expansion is $(1 + x)^{2n}$.

The coefficients of the second, third, and fourth terms are in Arithmetic Progression (A.P.).

To Show:

$2n^2 – 9n + 7 = 0$.

Proof:

The expansion is $(1+x)^{2n}$. The power of the binomial is $N=2n$.

The coefficient of the $m^{\text{th}}$ term in the expansion of $(1+x)^N$ is $\binom{N}{m-1}$.

Coefficient of the second term ($m=2$) is $\binom{2n}{2-1} = \binom{2n}{1}$.

Coefficient of the third term ($m=3$) is $\binom{2n}{3-1} = \binom{2n}{2}$.

Coefficient of the fourth term ($m=4$) is $\binom{2n}{4-1} = \binom{2n}{3}$.

We are given that these coefficients are in A.P.

If three numbers $a, b, c$ are in A.P., then $2b = a+c$.

So, $2 \cdot \binom{2n}{2} = \binom{2n}{1} + \binom{2n}{3}$.

Write out the binomial coefficients:

$\binom{2n}{1} = \frac{(2n)!}{1!(2n-1)!} = 2n$

$\binom{2n}{2} = \frac{(2n)!}{2!(2n-2)!} = \frac{2n(2n-1)}{2 \times 1} = n(2n-1)$

$\binom{2n}{3} = \frac{(2n)!}{3!(2n-3)!} = \frac{2n(2n-1)(2n-2)}{3 \times 2 \times 1} = \frac{2n(2n-1)2(n-1)}{6} = \frac{2n(2n-1)(n-1)}{3}$

Substitute these values into the A.P. condition:

We can assume $n \ge 2$ for the third and fourth terms to exist and for the denominators of the binomial coefficients to be non-zero. If $n=1$, the expansion is $(1+x)^2 = 1+2x+x^2$. The 2nd, 3rd, 4th terms are $2x, x^2, 0$. Coefficients are 2, 1, 0. Are 2, 1, 0 in A.P.? $2(1) = 2+0$, $2=2$. Yes, they are in A.P. If $n=1$, the equation $2n^2 - 9n + 7 = 0$ becomes $2(1)^2 - 9(1) + 7 = 2 - 9 + 7 = 0$. So $n=1$ satisfies the equation. Let's proceed assuming $n \ge 2$ and see if we get the same equation.

Divide equation (i) by $2n$ (since for terms to exist, $n \ge 2$, so $2n \neq 0$):

$2(2n-1) = 1 + \frac{(2n-1)(n-1)}{3}$

$4n - 2 = 1 + \frac{(2n-1)(n-1)}{3}$

Multiply by 3:

$3(4n - 2) = 3(1) + (2n-1)(n-1)$

$12n - 6 = 3 + (2n^2 - 2n - n + 1)$

$12n - 6 = 3 + 2n^2 - 3n + 1$

$12n - 6 = 2n^2 - 3n + 4$

Move all terms to one side to form a quadratic equation:

$0 = 2n^2 - 3n - 12n + 4 + 6$

$0 = 2n^2 - 15n + 10$

Wait, the required equation is $2n^2 - 9n + 7 = 0$. Let's recheck the A.P. condition and the calculation.

Coefficients are $c_1 = \binom{2n}{1}$, $c_2 = \binom{2n}{2}$, $c_3 = \binom{2n}{3}$.

$c_1 = 2n$

$c_2 = \frac{2n(2n-1)}{2} = n(2n-1)$

$c_3 = \frac{2n(2n-1)(2n-2)}{6} = \frac{2n(2n-1)2(n-1)}{6} = \frac{2n(2n-1)(n-1)}{3}$

A.P. condition: $2c_2 = c_1 + c_3$

$2 \cdot n(2n-1) = 2n + \frac{2n(2n-1)(n-1)}{3}$

Divide by $2n$ (assuming $n \ge 2$, so $2n \neq 0$):

$2n-1 = 1 + \frac{(2n-1)(n-1)}{3}$

Multiply by 3:

$3(2n-1) = 3 + (2n-1)(n-1)$

$6n - 3 = 3 + (2n^2 - 2n - n + 1)$

$6n - 3 = 3 + 2n^2 - 3n + 1$

$6n - 3 = 2n^2 - 3n + 4$

Move all terms to the right side:

$0 = 2n^2 - 3n - 6n + 4 + 3$

$0 = 2n^2 - 9n + 7$

This is the required equation.

We assumed $n \ge 2$. If $n=1$, the coefficients are $\binom{2}{1}=2$, $\binom{2}{2}=1$, $\binom{2}{3}=0$. These are 2, 1, 0, which are in A.P. The equation $2n^2 - 9n + 7 = 0$ for $n=1$ is $2(1)^2 - 9(1) + 7 = 2 - 9 + 7 = 0$, which is true. So $n=1$ is a possible value of $n$ that satisfies the condition.

The derivation $2n^2 - 9n + 7 = 0$ holds for $n \ge 2$. Since $n=1$ also satisfies the equation, the equation holds for all positive integers $n$ where the second, third, and fourth terms exist (i.e., $2n \ge 3$, which means $n \ge 3/2$). So, the derivation is valid for $n \ge 2$, and we checked $n=1$ separately.

We have shown that if the coefficients of the second, third, and fourth terms in the expansion of $(1+x)^{2n}$ are in A.P., then $2n^2 - 9n + 7 = 0$.

Given:

The expansion is $(1 + x + x^2 + x^3)^{11}$.

To Find:

The coefficient of $x^4$ in the expansion.

Solution:

Let the given expression be $E$.

$E = (1 + x + x^2 + x^3)^{11}$

We can factor the base expression $1 + x + x^2 + x^3$. This is a geometric series with first term 1, ratio $x$, and 4 terms.

$1 + x + x^2 + x^3 = \frac{1(1-x^4)}{1-x} = \frac{1-x^4}{1-x}$, provided $x \neq 1$.

Substitute this factorization back into the expression for $E$:

$E = \left( \frac{1-x^4}{1-x} \right)^{11} = \frac{(1-x^4)^{11}}{(1-x)^{11}}$

$E = (1-x^4)^{11} (1-x)^{-11}$

We need to find the coefficient of $x^4$ in the product of these two expansions.

Expand each factor using the binomial theorem:

$(1-x^4)^{11} = \binom{11}{0}(1)^{11}(-x^4)^0 + \binom{11}{1}(1)^{10}(-x^4)^1 + \binom{11}{2}(1)^{9}(-x^4)^2 + \binom{11}{3}(1)^{8}(-x^4)^3 + \dots$

$(1-x^4)^{11} = \binom{11}{0} - \binom{11}{1}x^4 + \binom{11}{2}x^8 - \binom{11}{3}x^{12} + \dots$

$= 1 - 11x^4 + \binom{11}{2}x^8 - \dots$

We are interested in terms up to $x^4$. So we only need the first two terms of this expansion:

$(1-x^4)^{11} = 1 - 11x^4 + (\text{terms with } x^8, x^{12}, \dots)$

Now consider the expansion of $(1-x)^{-11}$. We use the generalized binomial theorem for negative integer exponents: $(1-y)^{-n} = \sum_{k=0}^\infty \binom{n+k-1}{k} y^k$.

Here $y=x$ and $n=11$.

$(1-x)^{-11} = \sum_{k=0}^\infty \binom{11+k-1}{k} x^k = \sum_{k=0}^\infty \binom{10+k}{k} x^k$

We need terms up to $x^4$. So we need the terms for $k=0, 1, 2, 3, 4$:

$(1-x)^{-11} = \binom{10+0}{0}x^0 + \binom{10+1}{1}x^1 + \binom{10+2}{2}x^2 + \binom{10+3}{3}x^3 + \binom{10+4}{4}x^4 + \dots$

$(1-x)^{-11} = \binom{10}{0} + \binom{11}{1}x + \binom{12}{2}x^2 + \binom{13}{3}x^3 + \binom{14}{4}x^4 + \dots$

$= 1 + 11x + \frac{12 \times 11}{2}x^2 + \frac{13 \times 12 \times 11}{3 \times 2 \times 1}x^3 + \frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1}x^4 + \dots$

$= 1 + 11x + 66x^2 + 286x^3 + 1001x^4 + \dots$

Now, multiply the two expansions and collect the terms with $x^4$:

$E = (1 - 11x^4 + \dots) \cdot (1 + 11x + 66x^2 + 286x^3 + 1001x^4 + \dots)$

The terms that produce $x^4$ are formed by multiplying:

1. Constant term from $(1-x^4)^{11}$ with $x^4$ term from $(1-x)^{-11}$:

$1 \cdot (\binom{14}{4}x^4) = 1001x^4$

2. $x^4$ term from $(1-x^4)^{11}$ with constant term from $(1-x)^{-11}$:

$(-11x^4) \cdot (\binom{10}{0}) = -11x^4 \cdot 1 = -11x^4$

Any other combination of terms from the two limited expansions shown will result in a power of $x$ higher than 4. For example, the $x^8$ term from the first factor multiplied by any term from the second factor will give $x^8$ or higher.

The coefficient of $x^4$ is the sum of the coefficients of these terms:

Coefficient of $x^4 = 1001 + (-11) = 1001 - 11 = 990$.

The coefficient of $x^4$ in the expansion of $(1 + x + x^2 + x^3)^{11}$ is 990.

The final answer is $\boxed{990}$.

Given:

The expansion is $\left( \frac{p}{2} + 2 \right)^8$, where $p$ is a real number.

The middle term in the expansion is 1120.

To Find:

The value of $p$.

Solution:

The given expansion is of the form $(a+b)^n$, where $a = \frac{p}{2}$, $b = 2$, and $n = 8$.

The total number of terms in the expansion is $n+1 = 8+1 = 9$.

Since the number of terms (9) is odd, there is one middle term.

The position of the middle term is $\frac{n}{2} + 1$ when $n$ is even.

Middle term position = $\frac{8}{2} + 1 = 4 + 1 = 5^{\text{th}}$ term.

We need to find the 5^th term ($T_5$) of the expansion.

The general term, $T_{k+1}$, in the expansion of $(a+b)^n$ is given by $T_{k+1} = \binom{n}{k} a^{n-k} b^k$.

For the 5^th term, $k+1=5$, so $k=4$.

Substitute the values of $n=8$, $k=4$, $a=\frac{p}{2}$, and $b=2$ into the general term formula:

$T_5 = \binom{8}{4} \left(\frac{p}{2}\right)^{8-4} (2)^4$

$T_5 = \binom{8}{4} \left(\frac{p}{2}\right)^4 (2)^4$

Calculate the binomial coefficient $\binom{8}{4}$:

$\binom{8}{4} = \frac{8!}{4!4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = \frac{\cancel{8}^{1} \times 7 \times \cancel{6}^{1} \times 5}{\cancel{24}_{1}} = 7 \times 5 \times 1 = 70$

Calculate the powers of the terms:

$\left(\frac{p}{2}\right)^4 = \frac{p^4}{2^4} = \frac{p^4}{16}$

$(2)^4 = 16$

Substitute these values back into the expression for $T_5$:

$T_5 = 70 \cdot \frac{p^4}{16} \cdot 16$

Cancel out the 16:

$T_5 = 70 p^4$

We are given that the middle term ($T_5$) is 1120.

Divide both sides by 70:

$p^4 = \frac{1120}{70}$

$p^4 = \frac{112}{7}$

Divide 112 by 7:

$\begin{array}{r} 16 \\ 7{\overline{\smash{\big)}\,112}} \\ \underline{-~\phantom{(}(7)}\phantom{0} \\ 42\phantom{)} \\ \underline{-~\phantom{()}(42)} \\ 0\phantom{)} \end{array}$

So, $p^4 = 16$.

To find $p$, take the fourth root of 16:

$p = \pm \sqrt[4]{16}$

Since $2^4 = 16$ and $(-2)^4 = 16$, the possible real values for $p$ are 2 and -2.

We are given that $p$ is a real number, so both 2 and -2 are valid solutions.

The possible values of $p$ are 2 and -2.

The final answer is $\boxed{\pm 2}$.

Given:

The expansion is $\left( x − \frac{1}{x} \right)^{2n}$.

To Show:

The middle term is $\frac{1 \times 3 \times 5 \times \dots (2n − 1)}{n!} \times (-2)^n$.

Proof:

The given expansion is of the form $(a+b)^N$, where $a = x$, $b = -\frac{1}{x}$, and $N = 2n$.

The total number of terms in the expansion is $N+1 = 2n+1$.

Since the number of terms ($2n+1$) is odd, there is one middle term.

The position of the middle term is $\frac{N}{2} + 1$ when $N$ is even.

Middle term position = $\frac{2n}{2} + 1 = n + 1^{\text{th}}$ term.

We need to find the $(n+1)^{\text{th}}$ term ($T_{n+1}$) of the expansion.

The general term, $T_{k+1}$, in the expansion of $(a+b)^N$ is given by $T_{k+1} = \binom{N}{k} a^{N-k} b^k$.

For the $(n+1)^{\text{th}}$ term, $k+1 = n+1$, so $k=n$.

Substitute the values of $N=2n$, $k=n$, $a=x$, and $b=-\frac{1}{x}$ into the general term formula:

$T_{n+1} = \binom{2n}{n} (x)^{2n-n} \left(-\frac{1}{x}\right)^n$

$T_{n+1} = \binom{2n}{n} (x)^n \left(-\frac{1}{x}\right)^n$

$T_{n+1} = \binom{2n}{n} x^n (-1)^n \left(\frac{1}{x}\right)^n$

$T_{n+1} = \binom{2n}{n} x^n (-1)^n \frac{1}{x^n}$

Cancel $x^n$ (assuming $x \neq 0$):

$T_{n+1} = \binom{2n}{n} (-1)^n$

Now, we need to express $\binom{2n}{n}$ in the required form.

$\binom{2n}{n} = \frac{(2n)!}{n!n!}$

Expand $(2n)!$:

$(2n)! = (2n) \times (2n-1) \times (2n-2) \times (2n-3) \times \dots \times 3 \times 2 \times 1$

Group the even and odd terms:

$(2n)! = [(2n) \times (2n-2) \times \dots \times 2] \times [(2n-1) \times (2n-3) \times \dots \times 1]$

Factor out 2 from each term in the first bracket (there are $n$ terms):

$(2n) \times (2n-2) \times \dots \times 2 = (2 \times n) \times (2 \times (n-1)) \times \dots \times (2 \times 1)$

$= 2^n \times [n \times (n-1) \times \dots \times 1] = 2^n \cdot n!$

So, $(2n)! = (2^n \cdot n!) \times [1 \times 3 \times 5 \times \dots \times (2n-1)]$

Substitute this back into the expression for $\binom{2n}{n}$:

$\binom{2n}{n} = \frac{2^n \cdot n! \cdot [1 \times 3 \times 5 \times \dots \times (2n-1)]}{n!n!}$

Cancel one $n!$ from the numerator and denominator:

$\binom{2n}{n} = \frac{2^n \cdot [1 \times 3 \times 5 \times \dots \times (2n-1)]}{n!}$

Now, substitute this expression for $\binom{2n}{n}$ back into the middle term formula $T_{n+1} = \binom{2n}{n} (-1)^n$:

$T_{n+1} = \frac{2^n \cdot [1 \times 3 \times 5 \times \dots \times (2n-1)]}{n!} \cdot (-1)^n$

$T_{n+1} = \frac{[1 \times 3 \times 5 \times \dots \times (2n-1)]}{n!} \cdot 2^n \cdot (-1)^n$

$T_{n+1} = \frac{1 \times 3 \times 5 \times \dots (2n − 1)}{n!} \times (-2)^n$

This matches the required expression for the middle term.

The middle term in the expansion of $\left( x − \frac{1}{x} \right)^{2n}$ is $\frac{1 \times 3 \times 5 \times … (2n − 1)}{n!} \times (-2)^n$.

Given:

The binomial expansion is $\left( \sqrt[3]{2} + \frac{1}{\sqrt[3]{3}} \right)^n$.

The ratio of the 7^th term from the beginning to the 7^th term from the end is $\frac{1}{6}$.

To Find:

The value of $n$.

Solution:

The given expansion is of the form $(a+b)^n$, where $a = \sqrt[3]{2} = 2^{1/3}$ and $b = \frac{1}{\sqrt[3]{3}} = 3^{-1/3}$.

The general term, $T_{k+1}$, in the expansion of $(a+b)^n$ is given by $T_{k+1} = \binom{n}{k} a^{n-k} b^k$.

7^th term from the beginning ($T_7$):

For the 7^th term from the beginning, we set $k+1=7$, so $k=6$.

$T_7 = \binom{n}{6} (2^{1/3})^{n-6} (3^{-1/3})^6$

$T_7 = \binom{n}{6} 2^{(1/3)(n-6)} 3^{(-1/3) \times 6}$

$T_7 = \binom{n}{6} 2^{\frac{n-6}{3}} 3^{-2}$

7^th term from the end:

The total number of terms in the expansion is $n+1$. The $r^{\text{th}}$ term from the end is the $( (n+1) - r + 1 )^{\text{th}}$ term from the beginning.

For the 7^th term from the end, $r=7$. The position from the beginning is $(n+1 - 7 + 1) = (n-5)^{\text{th}}$.

The $(n-5)^{\text{th}}$ term from the beginning is $T_{n-5}$. For this term, $k+1 = n-5$, so $k=n-6$.

$T_{n-5} = \binom{n}{n-6} (2^{1/3})^{n-(n-6)} (3^{-1/3})^{n-6}$

$T_{n-5} = \binom{n}{n-6} (2^{1/3})^6 (3^{-1/3})^{n-6}$

$T_{n-5} = \binom{n}{n-6} 2^{6/3} 3^{(-1/3)(n-6)}$

$T_{n-5} = \binom{n}{n-6} 2^2 3^{-\frac{n-6}{3}}$

Using the property $\binom{n}{n-k} = \binom{n}{k}$, we have $\binom{n}{n-6} = \binom{n}{6}$.

$T_{n-5} = \binom{n}{6} 2^2 3^{-\frac{n-6}{3}}$

We are given that the ratio of the 7^th term from the beginning to the 7^th term from the end is $\frac{1}{6}$.

Substitute the expressions for $T_7$ and $T_{n-5}$ from (i) and (ii):

$\frac{\binom{n}{6} 2^{\frac{n-6}{3}} 3^{-2}}{\binom{n}{6} 2^2 3^{-\frac{n-6}{3}}} = \frac{1}{6}$

Cancel out $\binom{n}{6}$ (assuming $n \ge 6$ for the 7th term to exist) and simplify the powers using exponent rules:

$2^{\frac{n-6}{3} - 2} \cdot 3^{-2 - (-\frac{n-6}{3})} = \frac{1}{6}$

$2^{\frac{n-6-6}{3}} \cdot 3^{-2 + \frac{n-6}{3}} = 6^{-1}$

$2^{\frac{n-12}{3}} \cdot 3^{\frac{-6+n-6}{3}} = 6^{-1}$

$2^{\frac{n-12}{3}} \cdot 3^{\frac{n-12}{3}} = 6^{-1}$

$(2 \cdot 3)^{\frac{n-12}{3}} = 6^{-1}$

$6^{\frac{n-12}{3}} = 6^{-1}$

Since the bases are equal, the exponents must be equal:

$\frac{n-12}{3} = -1$

Multiply both sides by 3:

$n - 12 = -3$

Add 12 to both sides:

$n = -3 + 12$

$n = 9$

Check if $n=9$ is valid. For the 7th term to exist, $n$ must be at least 6. $9 \ge 6$, so this is valid.

The value of $n$ is 9.

The final answer is $\boxed{9}$.

Given:

The expansion is $(x+a)^n$.

$O$ = sum of odd terms in the expansion.

$E$ = sum of even terms in the expansion.

To Prove:

(i) $O^2 – E^2 = (x^2 – a^2)^n$

(ii) $4OE = (x + a)^{2n} – (x – a)^{2n}$

Proof:

The binomial expansion of $(x+a)^n$ is:

$(x+a)^n = \binom{n}{0}x^n a^0 + \binom{n}{1}x^{n-1} a^1 + \binom{n}{2}x^{n-2} a^2 + \binom{n}{3}x^{n-3} a^3 + \dots$

The terms in the expansion are $T_1, T_2, T_3, T_4, \dots$.

$T_1 = \binom{n}{0}x^n a^0$ (Odd term, as the index of the term is 1)

$T_2 = \binom{n}{1}x^{n-1} a^1$ (Even term, as the index of the term is 2)

$T_3 = \binom{n}{2}x^{n-2} a^2$ (Odd term)

$T_4 = \binom{n}{3}x^{n-3} a^3$ (Even term)

And so on.

The sum of odd terms, $O$, is the sum of $T_1, T_3, T_5, \dots$:

$O = \binom{n}{0}x^n a^0 + \binom{n}{2}x^{n-2} a^2 + \binom{n}{4}x^{n-4} a^4 + \dots$

The sum of even terms, $E$, is the sum of $T_2, T_4, T_6, \dots$:

$E = \binom{n}{1}x^{n-1} a^1 + \binom{n}{3}x^{n-3} a^3 + \binom{n}{5}x^{n-5} a^5 + \dots$

The sum of the entire expansion is $O+E$:

Now consider the expansion of $(x-a)^n$. Substitute $a$ with $-a$ in the expansion of $(x+a)^n$.

$(x-a)^n = \binom{n}{0}x^n (-a)^0 + \binom{n}{1}x^{n-1} (-a)^1 + \binom{n}{2}x^{n-2} (-a)^2 + \binom{n}{3}x^{n-3} (-a)^3 + \dots$

$(x-a)^n = \binom{n}{0}x^n a^0 - \binom{n}{1}x^{n-1} a^1 + \binom{n}{2}x^{n-2} a^2 - \binom{n}{3}x^{n-3} a^3 + \dots$

The terms with an even index for the power of $a$ (which correspond to odd terms in the $(x+a)^n$ expansion) remain positive. The terms with an odd index for the power of $a$ (which correspond to even terms in the $(x+a)^n$ expansion) become negative.

So, the expansion of $(x-a)^n$ is the sum of the odd terms of $(x+a)^n$ minus the sum of the even terms of $(x+a)^n$.

Part (i) Proof:

Consider the expression $O^2 - E^2$. This is a difference of squares, which can be factored as $(O-E)(O+E)$.

$O^2 - E^2 = (O-E)(O+E)$

Substitute the values of $(O+E)$ from (i) and $(O-E)$ from (ii):

$O^2 - E^2 = (x-a)^n \cdot (x+a)^n$

Using the property $(bc)^m = b^m c^m$, we have $(x-a)^n (x+a)^n = ((x-a)(x+a))^n$.

Using the difference of squares formula, $(x-a)(x+a) = x^2 - a^2$.

$O^2 - E^2 = (x^2 - a^2)^n$

Thus, the first part of the identity is proven.

Part (ii) Proof:

Consider the expression $4OE$. We can write this as $4 \cdot (O \cdot E)$.

We know $(O+E) = (x+a)^n$ and $(O-E) = (x-a)^n$.

Multiply these two equations:

$(O+E)(O-E) = (x+a)^n (x-a)^n$

$O^2 - E^2 = (x^2 - a^2)^n$ (This brings us back to part (i))

Let's consider $(O+E)^2$ and $(O-E)^2$.

$(O+E)^2 = ((x+a)^n)^2 = (x+a)^{2n}$

$(O-E)^2 = ((x-a)^n)^2 = (x-a)^{2n}$

Expand the left sides:

$(O+E)^2 = O^2 + 2OE + E^2$

$(O-E)^2 = O^2 - 2OE + E^2$

Subtract the second expanded equation from the first:

$(O^2 + 2OE + E^2) - (O^2 - 2OE + E^2) = (O+E)^2 - (O-E)^2$

$O^2 + 2OE + E^2 - O^2 + 2OE - E^2 = (O+E)^2 - (O-E)^2$

$4OE = (O+E)^2 - (O-E)^2$

Substitute the expressions for $(O+E)^2$ and $(O-E)^2$:

$4OE = (x+a)^{2n} - (x-a)^{2n}$

Thus, the second part of the identity is proven.

We have shown that $O^2 – E^2 = (x^2 – a^2)^n$ and $4OE = (x + a)^{2n} – (x – a)^{2n}$.

Given:

The expansion is $\left( x^2 + \frac{1}{x} \right)^{2n}$.

The term containing $x^p$ occurs in this expansion, where $p$ is an integer.

To Prove:

The coefficient of $x^p$ is $\frac{(2n)!}{\left(\frac{4n - p}{3}\right)! \left(\frac{2n + p}{3}\right)!}$.

Proof:

The given expansion is of the form $(a+b)^N$, where $a = x^2$, $b = \frac{1}{x} = x^{-1}$, and $N = 2n$.

The general term, $T_{k+1}$, in the expansion of $(a+b)^N$ is given by the formula:

$T_{k+1} = \binom{N}{k} a^{N-k} b^k$

Substitute the values of $N=2n$, $a=x^2$, and $b=x^{-1}$ for the given expansion:

$T_{k+1} = \binom{2n}{k} (x^2)^{2n-k} (x^{-1})^k$

Simplify the terms involving $x$:

$(x^2)^{2n-k} = x^{2(2n-k)} = x^{4n-2k}$

$(x^{-1})^k = x^{-k}$

The variable part of the general term is the product of the terms involving $x$:

$x^{4n-2k} \cdot x^{-k} = x^{(4n-2k) - k} = x^{4n-3k}$

The general term is $T_{k+1} = \binom{2n}{k} x^{4n-3k}$.

We are told that $x^p$ occurs in the expansion. This means there exists a value of $k$ (where $k$ is a non-negative integer and $0 \leq k \leq 2n$) such that the power of $x$ in the general term is equal to $p$.

Set the exponent of $x$ equal to $p$:

$4n - 3k = p$

Solve for $k$:

$4n - p = 3k$

$k = \frac{4n - p}{3}$

For a term containing $x^p$ to exist, $k$ must be a non-negative integer such that $0 \leq k \leq 2n$. This implies that $(4n-p)$ must be divisible by 3, and the value of $k$ must fall within the valid range for the binomial coefficient.

If such an integer $k$ exists, the coefficient of $x^p$ is given by $\binom{2n}{k}$ with $k = \frac{4n-p}{3}$.

Coefficient of $x^p = \binom{2n}{\frac{4n-p}{3}}$

Using the definition of the binomial coefficient $\binom{N}{K} = \frac{N!}{K!(N-K)!}$, we have:

Coefficient of $x^p = \frac{(2n)!}{\left(\frac{4n-p}{3}\right)! \left(2n - \frac{4n-p}{3}\right)!}$

Simplify the term in the second factorial in the denominator:

$2n - \frac{4n-p}{3} = \frac{3(2n) - (4n-p)}{3} = \frac{6n - 4n + p}{3} = \frac{2n + p}{3}$

Substitute this back into the denominator:

Coefficient of $x^p = \frac{(2n)!}{\left(\frac{4n-p}{3}\right)! \left(\frac{2n+p}{3}\right)!}$

This is the required expression for the coefficient of $x^p$. The existence of the term $x^p$ guarantees that $\frac{4n-p}{3}$ is a non-negative integer $\le 2n$.

Thus, if $x^p$ occurs in the expansion of $\left( x^2 + \frac{1}{x} \right)^{2n}$, its coefficient is $\frac{(2n)!}{\left(\frac{4n - p}{3}\right)! \left(\frac{2n + p}{3}\right)!}$.

Given:

The expression is $(1 + x + 2x^3) \left( \frac{3}{2} x^2 - \frac{1}{3x} \right)^9$.

To Find:

The term independent of $x$ in the expansion of the given expression.

Solution:

Let the given expression be $E$.

$E = (1 + x + 2x^3) \cdot \left( \frac{3}{2} x^2 - \frac{1}{3x} \right)^9$

Let the expansion of $\left( \frac{3}{2} x^2 - \frac{1}{3x} \right)^9$ be denoted by $S$. We need to find the terms in $S$ that, when multiplied by the terms in $(1 + x + 2x^3)$, result in a term independent of $x$ (i.e., a term with $x^0$).

The terms in the first factor are $1, x, 2x^3$.

To get a term independent of $x$ in the product, we need to multiply:

1. The constant term (term with $x^0$) from $(1 + x + 2x^3)$ by the term with $x^0$ from $S$.

2. The term with $x^1$ from $(1 + x + 2x^3)$ by the term with $x^{-1}$ from $S$.

3. The term with $x^3$ from $(1 + x + 2x^3)$ by the term with $x^{-3}$ from $S$.

Let's find the general term of the expansion $S = \left( \frac{3}{2} x^2 - \frac{1}{3x} \right)^9$. This is of the form $(a+b)^n$, where $a = \frac{3}{2} x^2$, $b = -\frac{1}{3x}$, and $n = 9$.

The general term, $T_{k+1}$, is $T_{k+1} = \binom{n}{k} a^{n-k} b^k$.

$T_{k+1} = \binom{9}{k} \left(\frac{3}{2} x^2\right)^{9-k} \left(-\frac{1}{3x}\right)^k$

Simplify the terms involving $x$ and the constant terms:

Constant part: $\binom{9}{k} \left(\frac{3}{2}\right)^{9-k} \left(-\frac{1}{3}\right)^k$

Variable part: $(x^2)^{9-k} \cdot \left(\frac{1}{x}\right)^k = x^{2(9-k)} \cdot x^{-k} = x^{18-2k} \cdot x^{-k} = x^{18-3k}$

The general term of $S$ is $T_{k+1} = \binom{9}{k} \left(\frac{3}{2}\right)^{9-k} \left(-\frac{1}{3}\right)^k x^{18-3k}$.

Now, let's find the required terms from the expansion $S$:

1. Term independent of $x$ (coefficient of $x^0$) in $S$: Set the exponent of $x$ to 0.

$18 - 3k = 0 \implies 3k = 18 \implies k = 6$.

This is the 7^th term ($T_7$) of $S$. The constant part is the term independent of $x$ in $S$.

Term with $x^0$ in $S = \binom{9}{6} \left(\frac{3}{2}\right)^{9-6} \left(-\frac{1}{3}\right)^6 = \binom{9}{6} \left(\frac{3}{2}\right)^3 \left(-\frac{1}{3}\right)^6$

$\binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.

$\left(\frac{3}{2}\right)^3 = \frac{27}{8}$.

$\left(-\frac{1}{3}\right)^6 = \frac{1}{3^6} = \frac{1}{729}$.

Term with $x^0$ in $S = 84 \cdot \frac{27}{8} \cdot \frac{1}{729} = \frac{84 \times 27}{8 \times 729} = \frac{\cancel{84}^{21} \times \cancel{27}^{1}}{\cancel{8}_{2} \times \cancel{729}_{27}} = \frac{21 \times 1}{2 \times 27} = \frac{\cancel{21}^{7}}{\cancel{54}_{18}} = \frac{7}{18}$.

This term will be multiplied by the constant term (1) from $(1 + x + 2x^3)$. Contribution to the term independent of $x$ is $1 \times \frac{7}{18} = \frac{7}{18}$.

2. Term with $x^{-1}$ in $S$: Set the exponent of $x$ to -1.

$18 - 3k = -1 \implies 3k = 18 + 1 \implies 3k = 19 \implies k = \frac{19}{3}$.

Since $k = \frac{19}{3}$ is not an integer, there is no term with $x^{-1}$ in the expansion of $S$. The coefficient of $x^{-1}$ in $S$ is 0.

This term would be multiplied by the $x^1$ term from $(1 + x + 2x^3)$. Contribution to the term independent of $x$ is $x \times 0 = 0$.

3. Term with $x^{-3}$ in $S$: Set the exponent of $x$ to -3.

$18 - 3k = -3 \implies 3k = 18 + 3 \implies 3k = 21 \implies k = 7$.

Since $k=7$ is an integer and $0 \leq 7 \leq 9$, this is a valid term (the 8^th term, $T_8$) of $S$. The constant part is:

Coefficient of $x^{-3}$ in $S = \binom{9}{7} \left(\frac{3}{2}\right)^{9-7} \left(-\frac{1}{3}\right)^7 = \binom{9}{7} \left(\frac{3}{2}\right)^2 \left(-\frac{1}{3}\right)^7$

$\binom{9}{7} = \binom{9}{9-7} = \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36$.

$\left(\frac{3}{2}\right)^2 = \frac{9}{4}$.

$\left(-\frac{1}{3}\right)^7 = (-1)^7 \frac{1}{3^7} = -\frac{1}{2187}$.

Coefficient of $x^{-3}$ in $S = 36 \cdot \frac{9}{4} \cdot \left(-\frac{1}{2187}\right) = \frac{\cancel{36}^{9} \times 9}{\cancel{4}_{1} \times 2187} \times (-1) = -\frac{9 \times 9}{2187} = -\frac{81}{2187}$.

Simplify the fraction: $2187 = 3^7$, $81 = 3^4$. $\frac{81}{2187} = \frac{3^4}{3^7} = 3^{4-7} = 3^{-3} = \frac{1}{27}$.

Coefficient of $x^{-3}$ in $S = -\frac{1}{27}$.

This term will be multiplied by the $2x^3$ term from $(1 + x + 2x^3)$. Contribution to the term independent of $x$ is $2x^3 \times (-\frac{1}{27} x^{-3}) = 2 \times (-\frac{1}{27}) x^{3-3} = -\frac{2}{27}$.

The term independent of $x$ in the overall expansion is the sum of the contributions from these three cases:

Term independent of $x = (\text{Contribution from } 1 \cdot x^0 \text{ in } S) + (\text{Contribution from } x \cdot x^{-1} \text{ in } S) + (\text{Contribution from } 2x^3 \cdot x^{-3} \text{ in } S)$

Term independent of $x = \frac{7}{18} + 0 + \left(-\frac{2}{27}\right)$

Term independent of $x = \frac{7}{18} - \frac{2}{27}$

Find a common denominator, which is LCM(18, 27). $18 = 2 \times 3^2$, $27 = 3^3$. LCM(18, 27) = $2 \times 3^3 = 2 \times 27 = 54$.

$\frac{7}{18} = \frac{7 \times 3}{18 \times 3} = \frac{21}{54}$

$\frac{2}{27} = \frac{2 \times 2}{27 \times 2} = \frac{4}{54}$

Term independent of $x = \frac{21}{54} - \frac{4}{54} = \frac{21 - 4}{54} = \frac{17}{54}$.

The term independent of $x$ in the expansion of $(1 + x + 2x^3) \left( \frac{3}{2} x^2 - \frac{1}{3x} \right)^9$ is $\frac{17}{54}$.

The final answer is $\boxed{\frac{17}{54}}$.

Given:

The expression $(x + a)^{100} + (x – a)^{100}$.

To Find:

The total number of terms after simplification.

Solution:

Let the given expression be $E$.

$E = (x + a)^{100} + (x – a)^{100}$

We use the binomial expansions for $(A+B)^n$ and $(A-B)^n$:

$(A+B)^n = \binom{n}{0}A^n B^0 + \binom{n}{1}A^{n-1}B^1 + \binom{n}{2}A^{n-2}B^2 + \dots + \binom{n}{n}A^0 B^n$

$(A-B)^n = \binom{n}{0}A^n B^0 - \binom{n}{1}A^{n-1}B^1 + \binom{n}{2}A^{n-2}B^2 - \dots + (-1)^n \binom{n}{n}A^0 B^n$

Adding the two expansions:

$(A+B)^n + (A-B)^n = 2 \left[ \binom{n}{0}A^n B^0 + \binom{n}{2}A^{n-2}B^2 + \binom{n}{4}A^{n-4}B^4 + \dots \right]$

The terms that remain are those with even indices in the binomial coefficient (starting from index 0), multiplied by 2.

In our case, $A=x$, $B=a$, and $n=100$. So the expression is:

$(x+a)^{100} + (x-a)^{100} = 2 \left[ \binom{100}{0}x^{100}a^0 + \binom{100}{2}x^{98}a^2 + \binom{100}{4}x^{96}a^4 + \dots + \binom{100}{100}x^0a^{100} \right]$

$(x+a)^{100} + (x-a)^{100} = 2 \left[ \binom{100}{0}x^{100} + \binom{100}{2}x^{98}a^2 + \binom{100}{4}x^{96}a^4 + \dots + \binom{100}{100}a^{100} \right]$

The powers of $a$ in the remaining terms are $0, 2, 4, \dots, 100$. Since these powers of $a$ are all distinct even numbers, the terms $x^{100-k}a^k$ for $k \in \{0, 2, 4, \dots, 100\}$ are distinct terms and cannot be combined further by collecting like terms (assuming $x \neq 0$ and $a \neq 0$).

The number of terms in the simplified expansion is equal to the number of even integers from 0 to 100, inclusive.

To find the number of even integers from 0 to 100, we can list them or use a formula. The numbers are $0, 2, 4, \dots, 100$. Let the position be $m$, such that the $m^{\text{th}}$ term is $2(m-1)$. $100 = 2(m-1) \implies 50 = m-1 \implies m=51$. Alternatively, the terms are of the form $2k$, where $k=0, 1, 2, \dots, 50$. The number of values of $k$ is $50 - 0 + 1 = 51$.

Thus, there are 51 terms in the simplified expansion.

The total number of terms in the expansion of $(x + a)^{100} + (x – a)^{100}$ after simplification is 51.

The correct option is (C) 51.

Given:

Integers $r > 1$ and $n > 2$.

The expansion is $(1 + x)^{2n}$.

The coefficients of the $(3r)^{\text{th}}$ term and the $(r + 2)^{\text{nd}}$ term are equal.

To Find:

The relationship between $n$ and $r$ from the given options.

Solution:

The expansion is $(1+x)^{2n}$, so the power of the binomial is $N=2n$.

The coefficient of the $m^{\text{th}}$ term in the expansion of $(1+x)^N$ is $\binom{N}{m-1}$.

The coefficient of the $(3r)^{\text{th}}$ term is $\binom{2n}{(3r)-1} = \binom{2n}{3r-1}$.

The coefficient of the $(r+2)^{\text{nd}}$ term is $\binom{2n}{(r+2)-1} = \binom{2n}{r+1}$.

We are given that these coefficients are equal:

For the binomial coefficients $\binom{N}{a}$ and $\binom{N}{b}$ to be equal, there are two possibilities:

1. $a = b$

2. $a + b = N$

In our case, $N=2n$, $a=3r-1$, and $b=r+1$.

Case 1: $3r - 1 = r + 1$

$3r - r = 1 + 1$

$2r = 2$

$r = 1$

However, the problem states that $r > 1$. Thus, this case is not possible under the given conditions.

Case 2: $(3r - 1) + (r + 1) = 2n$

$3r - 1 + r + 1 = 2n$

$4r = 2n$

Divide both sides by 2:

$2r = n$

$n = 2r$

We must check if this relationship is consistent with the given conditions $r > 1$ and $n > 2$.

If $r > 1$, then $2r > 2$. Since $n=2r$, we have $n > 2$. This is consistent with the condition $n > 2$.

Also, for the binomial coefficients to be well-defined, the indices must be non-negative integers less than or equal to $N=2n$.

$3r-1 \ge 0 \implies 3r \ge 1 \implies r \ge 1/3$. Given $r>1$, this is satisfied.

$r+1 \ge 0 \implies r \ge -1$. Given $r>1$, this is satisfied.

$3r-1 \le 2n$. Substitute $n=2r$: $3r-1 \le 2(2r) \implies 3r-1 \le 4r \implies -1 \le r$. Given $r>1$, this is satisfied.

$r+1 \le 2n$. Substitute $n=2r$: $r+1 \le 2(2r) \implies r+1 \le 4r \implies 1 \le 3r \implies r \ge 1/3$. Given $r>1$, this is satisfied.

The derived relationship $n=2r$ is consistent with all given conditions.

The relationship between $n$ and $r$ is $n=2r$.

The correct option is (A) n = 2r.

Given:

The expansion is $(1 + x)^{24}$.

The ratio of the coefficients of two successive terms is 1:4.

To Find:

The positions of these two successive terms.

Solution:

The expansion is $(1+x)^{24}$, so $n=24$.

The coefficient of the $(k+1)^{\text{th}}$ term in the expansion of $(1+x)^n$ is $\binom{n}{k}$.

Let the two successive terms be the $(r+1)^{\text{th}}$ term and the $(r+2)^{\text{th}}$ term.

The coefficient of the $(r+1)^{\text{th}}$ term is $\binom{24}{r}$.

The coefficient of the $(r+2)^{\text{th}}$ term is $\binom{24}{r+1}$.

We are given that the ratio of their coefficients is 1:4.

There are two possibilities for the order of the ratio:

Possibility 1: $\frac{\text{Coefficient of }(r+1)^{\text{th}} \text{ term}}{\text{Coefficient of }(r+2)^{\text{th}} \text{ term}} = \frac{1}{4}$

Possibility 2: $\frac{\text{Coefficient of }(r+2)^{\text{th}} \text{ term}}{\text{Coefficient of }(r+1)^{\text{th}} \text{ term}} = \frac{1}{4}$

We know the property of binomial coefficients: $\frac{\binom{n}{k}}{\binom{n}{k+1}} = \frac{k+1}{n-k}$.

Using this property for Possibility 1, with $n=24$ and $k=r$:

$\frac{\binom{24}{r}}{\binom{24}{r+1}} = \frac{r+1}{24-r}$

So, $\frac{r+1}{24-r} = \frac{1}{4}$

Cross-multiply:

$4(r+1) = 1(24-r)$

$4r + 4 = 24 - r$

$4r + r = 24 - 4$

$5r = 20$

$r = \frac{20}{5}$

$r = 4$

If $r=4$, the successive terms are the $(r+1)^{\text{th}} = 5^{\text{th}}$ term and the $(r+2)^{\text{th}} = 6^{\text{th}}$ term.

Their coefficients are $\binom{24}{4}$ and $\binom{24}{5}$. The ratio $\frac{\binom{24}{4}}{\binom{24}{5}} = \frac{4+1}{24-4} = \frac{5}{20} = \frac{1}{4}$. This matches the given ratio 1:4.

Let's check Possibility 2, where the ratio is reversed:

$\frac{\binom{24}{r+1}}{\binom{24}{r}} = \frac{24-(r+1)+1}{r+1} = \frac{24-r-1+1}{r+1} = \frac{24-r}{r+1}$

So, $\frac{24-r}{r+1} = \frac{1}{4}$

Cross-multiply:

$4(24-r) = 1(r+1)$

$96 - 4r = r + 1$

$96 - 1 = r + 4r$

$95 = 5r$

$r = \frac{95}{5}$

$r = 19$

If $r=19$, the successive terms are the $(r+1)^{\text{th}} = 20^{\text{th}}$ term and the $(r+2)^{\text{th}} = 21^{\text{st}}$ term.

Their coefficients are $\binom{24}{19}$ and $\binom{24}{20}$. The ratio $\frac{\binom{24}{19}}{\binom{24}{20}} = \frac{19+1}{24-19} = \frac{20}{5} = 4$. The ratio $\frac{\binom{24}{20}}{\binom{24}{19}} = \frac{5}{20} = \frac{1}{4}$.

So, if the ratio of the 20^th term's coefficient to the 19^th term's coefficient is 1:4, then $r=19$. If the ratio of the 19^th term's coefficient to the 20^th term's coefficient is 1:4, then $r=4$.

The phrasing "two successive terms ... whose coefficients are in the ratio 1:4" usually implies the first term mentioned in the ratio is the earlier term in the expansion and the second term mentioned is the later term. So the $(r+1)^{\text{th}}$ term's coefficient has the ratio 1 to the $(r+2)^{\text{th}}$ term's coefficient which has the ratio 4.

Coefficient of $(r+1)^{\text{th}}$ : Coefficient of $(r+2)^{\text{th}} = 1 : 4$.

This corresponds to the case where $r=4$. The terms are the 5^th and 6^th terms.

The two successive terms are the 5^th and 6^th terms.

The correct option is (C) 5^th and 6^th.

Given:

Two expansions: $(1 + x)^{2n}$ and $(1 + x)^{2n – 1}$.

To Find:

The ratio of the coefficient of $x^n$ in the first expansion to the coefficient of $x^n$ in the second expansion.

Solution:

The coefficient of $x^k$ in the expansion of $(1+x)^N$ is $\binom{N}{k}$.

For the expansion $(1+x)^{2n}$, the power is $N=2n$. The coefficient of $x^n$ is obtained by setting $k=n$.

Coefficient of $x^n$ in $(1+x)^{2n}$ is $\binom{2n}{n}$.

For the expansion $(1+x)^{2n-1}$, the power is $N=2n-1$. The coefficient of $x^n$ is obtained by setting $k=n$.

Coefficient of $x^n$ in $(1+x)^{2n-1}$ is $\binom{2n-1}{n}$.

We need to find the ratio of these two coefficients: $\frac{\binom{2n}{n}}{\binom{2n-1}{n}}$.

Write out the binomial coefficients using factorials:

$\binom{2n}{n} = \frac{(2n)!}{n!(2n-n)!} = \frac{(2n)!}{n!n!}$

$\binom{2n-1}{n} = \frac{(2n-1)!}{n!((2n-1)-n)!} = \frac{(2n-1)!}{n!(n-1)!}$

Now, form the ratio:

Ratio $= \frac{\frac{(2n)!}{n!n!}}{\frac{(2n-1)!}{n!(n-1)!}}$

Ratio $= \frac{(2n)!}{n!n!} \times \frac{n!(n-1)!}{(2n-1)!}$

Ratio $= \frac{(2n)!}{(2n-1)!} \times \frac{n!(n-1)!}{n!n!}$

Simplify the factorial terms:

$\frac{(2n)!}{(2n-1)!} = \frac{2n \times (2n-1)!}{(2n-1)!} = 2n$

$\frac{n!(n-1)!}{n!n!} = \frac{n!}{n!} \times \frac{(n-1)!}{n!} = 1 \times \frac{(n-1)!}{n \times (n-1)!} = \frac{1}{n}$

Substitute these simplified terms back into the ratio:

Ratio $= 2n \times \frac{1}{n}$

Ratio $= \frac{2n}{n} = 2$

The ratio is 2:1.

The ratio of the coefficient of $x^n$ in the expansion of $(1 + x)^{2n}$ and $(1 + x)^{2n – 1}$ is 2:1.

The correct option is (D) 2 : 1.

Given: Coefficients of 2nd, 3rd, 4th terms of $(1+x)^n$ are in A.P.

To Find: Value of $n$.

Solution:

The coefficients of the 2nd, 3rd, and 4th terms are $\binom{n}{1}$, $\binom{n}{2}$, and $\binom{n}{3}$ respectively.

Since they are in A.P., we have:

$2 \binom{n}{2} = \binom{n}{1} + \binom{n}{3}$

Substitute the binomial coefficient formulas:

$2 \cdot \frac{n(n-1)}{2} = n + \frac{n(n-1)(n-2)}{6}$

$n(n-1) = n + \frac{n(n-1)(n-2)}{6}$

Assuming $n \ge 3$ (for the 4th term to exist), we can divide by $n$:

$n-1 = 1 + \frac{(n-1)(n-2)}{6}$

$6(n-1) = 6 + (n-1)(n-2)$

$6n - 6 = 6 + (n^2 - 3n + 2)$

$6n - 6 = n^2 - 3n + 8$

$n^2 - 9n + 14 = 0$

Factor the quadratic equation:

$(n-2)(n-7) = 0$

The possible values for $n$ are $n=2$ or $n=7$.

If $n=2$, the coefficients are $\binom{2}{1}=2$, $\binom{2}{2}=1$, $\binom{2}{3}=0$. These are 2, 1, 0, which are in A.P.

If $n=7$, the coefficients are $\binom{7}{1}=7$, $\binom{7}{2}=21$, $\binom{7}{3}=35$. These are 7, 21, 35, which are in A.P. ($2 \times 21 = 7 + 35$).

Both $n=2$ and $n=7$ are valid solutions. Among the given options, 7 is listed.

The value of $n$ is 7.

The correct option is (B) 7.

Given:

$A$ = coefficient of $x^n$ in $(1 + x)^{2n}$.

$B$ = coefficient of $x^n$ in $(1 + x)^{2n-1}$.

To Find:

The value of $\frac{A}{B}$.

Solution:

The coefficient of $x^k$ in the expansion of $(1+x)^N$ is $\binom{N}{k}$.

For the expansion $(1+x)^{2n}$, $N=2n$. The coefficient of $x^n$ is $A = \binom{2n}{n}$.

For the expansion $(1+x)^{2n-1}$, $N=2n-1$. The coefficient of $x^n$ is $B = \binom{2n-1}{n}$.

We need to find the ratio $\frac{A}{B} = \frac{\binom{2n}{n}}{\binom{2n-1}{n}}$.

Write out the binomial coefficients using factorials:

$A = \binom{2n}{n} = \frac{(2n)!}{n!(2n-n)!} = \frac{(2n)!}{n!n!}$

$B = \binom{2n-1}{n} = \frac{(2n-1)!}{n!((2n-1)-n)!} = \frac{(2n-1)!}{n!(n-1)!}$

Now, form the ratio $\frac{A}{B}$:

$\frac{A}{B} = \frac{\frac{(2n)!}{n!n!}}{\frac{(2n-1)!}{n!(n-1)!}}$

$\frac{A}{B} = \frac{(2n)!}{n!n!} \times \frac{n!(n-1)!}{(2n-1)!}$

$\frac{A}{B} = \frac{(2n)!}{(2n-1)!} \times \frac{n!(n-1)!}{n!n!}$

Simplify the factorial terms:

$\frac{(2n)!}{(2n-1)!} = \frac{2n \times (2n-1)!}{(2n-1)!} = 2n$

$\frac{n!(n-1)!}{n!n!} = \frac{n!}{n!} \times \frac{(n-1)!}{n!} = 1 \times \frac{(n-1)!}{n \times (n-1)!} = \frac{1}{n}$

Substitute these simplified terms back into the ratio $\frac{A}{B}$:

$\frac{A}{B} = 2n \times \frac{1}{n}$

$\frac{A}{B} = \frac{2n}{n} = 2$

The value of $\frac{A}{B}$ is 2.

The correct option is (B) 2.

Given:

The expansion is $\left( \frac{1}{x} + x \sin x \right)^{10}$.

The value of the middle term is $7\frac{7}{8}$.

To Find:

The value of $x$.

Solution:

The given expansion is of the form $(a+b)^n$, where $a = \frac{1}{x}$, $b = x \sin x$, and $n = 10$.

The total number of terms in the expansion is $n+1 = 10+1 = 11$.

Since the number of terms (11) is odd, there is one middle term.

The position of the middle term is $\frac{n}{2} + 1$ when $n$ is even.

Middle term position = $\frac{10}{2} + 1 = 5 + 1 = 6^{\text{th}}$ term.

We need to find the 6^th term ($T_6$) of the expansion.

The general term, $T_{k+1}$, in the expansion of $(a+b)^n$ is given by $T_{k+1} = \binom{n}{k} a^{n-k} b^k$.

For the 6^th term, $k+1=6$, so $k=5$.

Substitute the values of $n=10$, $k=5$, $a=\frac{1}{x}$, and $b=x \sin x$ into the general term formula:

$T_6 = \binom{10}{5} \left(\frac{1}{x}\right)^{10-5} (x \sin x)^5$

$T_6 = \binom{10}{5} \left(\frac{1}{x}\right)^5 x^5 (\sin x)^5$

$T_6 = \binom{10}{5} \frac{1}{x^5} x^5 \sin^5 x$

Assuming $x \neq 0$, we can cancel $x^5$:

$T_6 = \binom{10}{5} \sin^5 x$

Calculate the binomial coefficient $\binom{10}{5}$:

$\binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$

So, $T_6 = 252 \sin^5 x$.

We are given that the middle term ($T_6$) is equal to $7\frac{7}{8}$. Convert this mixed number to an improper fraction:

$7\frac{7}{8} = \frac{7 \times 8 + 7}{8} = \frac{56 + 7}{8} = \frac{63}{8}$

Set the expression for $T_6$ equal to the given value:

Solve for $\sin^5 x$:

$\sin^5 x = \frac{63}{8 \times 252}$

Simplify the fraction. Notice that $252 = 4 \times 63$.

$\sin^5 x = \frac{\cancel{63}}{8 \times 4 \times \cancel{63}} = \frac{1}{8 \times 4} = \frac{1}{32}$

We have $\sin^5 x = \frac{1}{32}$. Take the 5^th root of both sides:

$\sqrt[5]{\sin^5 x} = \sqrt[5]{\frac{1}{32}}$

$\sin x = \frac{\sqrt[5]{1}}{\sqrt[5]{32}} = \frac{1}{2}$

Now we need to find the general solution for the trigonometric equation $\sin x = \frac{1}{2}$.

The principal value of $x$ for which $\sin x = \frac{1}{2}$ in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ is $\frac{\pi}{6}$.

The general solution for $\sin x = \sin \alpha$ is given by $x = n\pi + (-1)^n \alpha$, where $n$ is an integer ($n \in \mathbb{Z}$).

Using $\alpha = \frac{\pi}{6}$, the general solution for $\sin x = \frac{1}{2}$ is:

$x = n\pi + (-1)^n \frac{\pi}{6}$, where $n \in \mathbb{Z}$.

Comparing this general solution with the given options, we see that Option (C) matches this form.

The value of $x$ is $n\pi + (-1)^n \frac{\pi}{6}$, where $n$ is an integer.

The correct option is (C) $nπ + (-1)^n \frac{π}{6}$.

Given:

The expansion is $(1 + x)^{30}$.

To Find:

The largest coefficient in the expansion.

Solution:

The expansion is of the form $(1+x)^n$, where $n=30$.

The coefficients in the expansion of $(1+x)^n$ are $\binom{n}{k}$ for $k=0, 1, \dots, n$.

The binomial coefficients $\binom{n}{k}$ are symmetric, i.e., $\binom{n}{k} = \binom{n}{n-k}$.

When $n$ is even, the largest coefficient occurs at the middle term(s). In the case of $(1+x)^n$ with $n$ even, there is one middle term. The middle index is $\frac{n}{2}$.

Here, $n=30$, which is even. The middle index is $\frac{30}{2} = 15$.

The largest coefficient is $\binom{30}{15}$.

The largest coefficient in the expansion of $(1 + x)^{30}$ is $\binom{30}{15}$.

The final answer is $\boxed{\binom{30}{15}}$.

Given:

The expansion is $(x + y + z)^n$.

To Find:

The number of terms in the expansion.

Solution:

The expansion of a multinomial $(a_1 + a_2 + \dots + a_m)^n$ has terms of the form $C \cdot a_1^{n_1} a_2^{n_2} \dots a_m^{n_m}$, where $n_1, n_2, \dots, n_m$ are non-negative integers such that $n_1 + n_2 + \dots + n_m = n$, and $C$ is the multinomial coefficient.

Each distinct combination of exponents $(n_1, n_2, \dots, n_m)$ corresponds to a distinct term in the expansion.

In this case, we have a trinomial $(x+y+z)^n$, so $m=3$. The terms are of the form $C \cdot x^{n_1} y^{n_2} z^{n_3}$, where $n_1, n_2, n_3$ are non-negative integers such that $n_1 + n_2 + n_3 = n$.

The number of distinct terms is equal to the number of distinct non-negative integer solutions to the equation $n_1 + n_2 + n_3 = n$.

This can be solved using stars and bars. We have $n$ "stars" to distribute among 3 "bins" (for $n_1, n_2, n_3$), requiring $m-1 = 3-1 = 2$ "bars". The number of arrangements of $n$ stars and 2 bars is the number of ways to choose the positions of the 2 bars out of $n+2$ total positions, which is $\binom{n+2}{2}$.

Number of terms $= \binom{n+2}{2} = \frac{(n+2)!}{2!(n+2-2)!} = \frac{(n+2)!}{2!n!} = \frac{(n+2)(n+1)\cancel{n!}}{2 \times 1 \times \cancel{n!}} = \frac{(n+1)(n+2)}{2}$.

The number of terms in the expansion of $(x + y + z)^n$ is $\frac{(n+1)(n+2)}{2}$.

The final answer is $\boxed{\frac{(n+1)(n+2)}{2}}$.

Given: Expansion $\left( x^2 - \frac{1}{x^2} \right)^{16}$.

To Find: Constant term (independent of $x$).

Solution:

The general term $T_{k+1}$ in the expansion of $(a+b)^n$ is $\binom{n}{k} a^{n-k} b^k$.

Here $a=x^2$, $b=-\frac{1}{x^2}= -x^{-2}$, $n=16$.

$T_{k+1} = \binom{16}{k} (x^2)^{16-k} (-x^{-2})^k = \binom{16}{k} x^{32-2k} (-1)^k x^{-2k} = \binom{16}{k} (-1)^k x^{32-4k}$.

For the term independent of $x$, the power of $x$ must be 0.

$32 - 4k = 0 \implies 4k = 32 \implies k = 8$.

The constant term is the value of $T_{k+1}$ when $k=8$ (excluding $x^0$).

Constant term $= \binom{16}{8} (-1)^8 = \binom{16}{8} \times 1 = \binom{16}{8}$.

$\binom{16}{8} = \frac{16!}{8!8!} = \frac{16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9}{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$

$\binom{16}{8} = \frac{\cancel{16}^2 \times \cancel{15}^1 \times \cancel{14}^1 \times 13 \times \cancel{12}^1 \times 11 \times \cancel{10}^1 \times 9}{\cancel{8}_1 \times \cancel{7}_1 \times \cancel{6}_1 \times \cancel{5}_1 \times \cancel{4}_1 \times \cancel{3}_1 \times \cancel{2}_1 \times 1}$ (Careful cancellation: $8 \times 2 = 16$, $5 \times 3 = 15$, $7 \times 1 = 7$, $6$, $4$, $1$ remain)

$\frac{16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9}{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = \frac{(8 \times 2) \times (5 \times 3) \times (7 \times 2) \times 13 \times (6 \times 2) \times 11 \times (5 \times 2) \times (3 \times 3)}{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$

Cancel common factors: $\frac{\cancel{8} \times \cancel{2} \times \cancel{5} \times \cancel{3} \times \cancel{7} \times \cancel{2} \times 13 \times \cancel{6} \times \cancel{2} \times 11 \times \cancel{5} \times \cancel{2} \times \cancel{3} \times 3}{\cancel{8} \times \cancel{7} \times \cancel{6} \times \cancel{5} \times \cancel{4} \times \cancel{3} \times \cancel{2} \times 1}$

= $\frac{2 \times 13 \times 2 \times 11 \times 2 \times 5 \times 3}{4 \times 1}$ (Using $16/(8 \times 2) \rightarrow 1$, $15/(5 \times 3) \rightarrow 1$, $14/7 \rightarrow 2$, $12/6 \rightarrow 2$, $10/(?)$, $9/(?)$ )

Let's do the calculation $13 \times 11 \times \frac{16 \times 15 \times 14 \times 12 \times 10 \times 9}{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$.

$\binom{16}{8} = \frac{16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9}{40320} = 12870$.

The value of the constant term is 12870.

The final answer is $\boxed{12870}$.

Given:

The expansion is $\left( \sqrt[3]{2} + \frac{1}{\sqrt[3]{3}} \right)^n$.

The seventh term from the beginning and the seventh term from the end are equal.

To Find:

The value of $n$.

Solution:

The given expansion is of the form $(a+b)^n$, where $a = \sqrt[3]{2} = 2^{1/3}$ and $b = \frac{1}{\sqrt[3]{3}} = 3^{-1/3}$.

The general term, $T_{k+1}$, in the expansion of $(a+b)^n$ is given by $T_{k+1} = \binom{n}{k} a^{n-k} b^k$.

Seventh term from the beginning ($T_7$):

For the 7^th term from the beginning, set $k+1=7$, so $k=6$.

$T_7 = \binom{n}{6} (2^{1/3})^{n-6} (3^{-1/3})^6 = \binom{n}{6} 2^{(n-6)/3} 3^{-6/3} = \binom{n}{6} 2^{(n-6)/3} 3^{-2}$.

Seventh term from the end:

The total number of terms is $n+1$. The $r^{\text{th}}$ term from the end is the $( (n+1) - r + 1 )^{\text{th}}$ term from the beginning. For $r=7$, the term is the $(n+1 - 7 + 1) = (n-5)^{\text{th}}$ term from the beginning.

The $(n-5)^{\text{th}}$ term from the beginning is $T_{n-5}$. For this term, $k+1 = n-5$, so $k=n-6$.

$T_{n-5} = \binom{n}{n-6} (2^{1/3})^{n-(n-6)} (3^{-1/3})^{n-6} = \binom{n}{n-6} (2^{1/3})^6 (3^{-1/3})^{n-6} = \binom{n}{n-6} 2^{6/3} 3^{-(n-6)/3} = \binom{n}{n-6} 2^2 3^{-(n-6)/3}$.

Using $\binom{n}{n-6} = \binom{n}{6}$, we have $T_{n-5} = \binom{n}{6} 2^2 3^{-(n-6)/3}$.

We are given that the seventh term from the beginning and the seventh term from the end are equal:

Substitute the expressions for $T_7$ and $T_{n-5}$ (assuming $n \ge 6$ for the 7th term to exist):

$\binom{n}{6} 2^{(n-6)/3} 3^{-2} = \binom{n}{6} 2^2 3^{-(n-6)/3}$

Since $n \ge 6$, $\binom{n}{6} \neq 0$, so we can divide both sides by $\binom{n}{6}$:

$2^{(n-6)/3} 3^{-2} = 2^2 3^{-(n-6)/3}$

Rearrange the terms to group powers of the same base:

$\frac{2^{(n-6)/3}}{2^2} = \frac{3^{-(n-6)/3}}{3^{-2}}$

$2^{(n-6)/3 - 2} = 3^{-(n-6)/3 - (-2)}$

$2^{\frac{n-6-6}{3}} = 3^{\frac{-(n-6)+6}{3}}$

$2^{\frac{n-12}{3}} = 3^{\frac{-n+6+6}{3}}$

$2^{\frac{n-12}{3}} = 3^{\frac{12-n}{3}}$

$2^{\frac{n-12}{3}} = 3^{-\frac{n-12}{3}}$

$2^{\frac{n-12}{3}} = \frac{1}{3^{\frac{n-12}{3}}}$

$2^{\frac{n-12}{3}} \cdot 3^{\frac{n-12}{3}} = 1$

$(2 \cdot 3)^{\frac{n-12}{3}} = 1$

$6^{\frac{n-12}{3}} = 1$

For an exponential term with a base other than 1 or -1 to be equal to 1, the exponent must be 0.

So, $\frac{n-12}{3} = 0$

$n - 12 = 0$

$n = 12$

Check if $n=12$ is valid. For the 7th term to exist, $n \ge 6$. $12 \ge 6$, so this is valid.

The value of $n$ is 12.

The final answer is $\boxed{12}$.

Given:

The expansion is $\left( \frac{1}{a} − \frac{2b}{3} \right)^{10}$.

To Find:

The coefficient of $a^{-6} b^4$ in the expansion.

Solution:

The given expansion is of the form $(x+y)^n$, where $x = \frac{1}{a} = a^{-1}$, $y = -\frac{2b}{3} = -\frac{2}{3} b$, and $n = 10$.

The general term, $T_{k+1}$, in the expansion of $(x+y)^n$ is given by the formula:

$T_{k+1} = \binom{n}{k} x^{n-k} y^k$

Substitute the values of $n=10$, $x=a^{-1}$, and $y=-\frac{2}{3}b$ for the given expansion:

$T_{k+1} = \binom{10}{k} (a^{-1})^{10-k} \left(-\frac{2}{3} b\right)^k$

Simplify the terms involving $a$ and $b$, and the constant terms separately.

Constant part: $\binom{10}{k} \left(-\frac{2}{3}\right)^k$

Variable part: $(a^{-1})^{10-k} \cdot (b)^k = a^{-(10-k)} \cdot b^k = a^{-10+k} b^k$

The general term is $T_{k+1} = \binom{10}{k} \left(-\frac{2}{3}\right)^k a^{-10+k} b^k$.

We want to find the coefficient of $a^{-6} b^4$. By comparing the exponents of $a$ and $b$ in the general term with the target term $a^{-6} b^4$, we can find the value of $k$.

Exponent of $b$: $k = 4$

Let's check if this value of $k$ gives the correct exponent for $a$:

Exponent of $a$: $-10 + k = -10 + 4 = -6$.

This matches the required exponent of $a$. Since $k=4$ is a non-negative integer and $0 \leq 4 \leq 10$, this value of $k$ is valid.

The coefficient of $a^{-6} b^4$ is the constant part of $T_{k+1}$ when $k=4$.

Coefficient of $a^{-6} b^4 = \binom{10}{4} \left(-\frac{2}{3}\right)^4$

Calculate the binomial coefficient $\binom{10}{4}$:

$\binom{10}{4} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \frac{\cancel{10}^{5} \times \cancel{9}^{3} \times \cancel{8}^{1} \times 7}{\cancel{24}_{1}} = 5 \times 3 \times 1 \times 7 = 210$

Calculate $\left(-\frac{2}{3}\right)^4$:

$\left(-\frac{2}{3}\right)^4 = (-1)^4 \frac{2^4}{3^4} = 1 \cdot \frac{16}{81} = \frac{16}{81}$

Multiply the results:

Coefficient of $a^{-6} b^4 = 210 \times \frac{16}{81}$

Simplify the fraction. 210 is divisible by 3 (sum of digits = 3), 81 is divisible by 9 (sum of digits = 9). $210/3 = 70$, $81/3 = 27$.

Coefficient $= \frac{210}{81} \times 16 = \frac{\cancel{210}^{70}}{\cancel{81}_{27}} \times 16 = \frac{70 \times 16}{27}$

$70 \times 16 = 1120$.

Coefficient $= \frac{1120}{27}$.

The term $T_5$ in the hint is the general term $T_{k+1}$ with $k=4$. The hint provides the term as $\frac{1120}{27} a^{-6} b^4$, confirming our calculation of the coefficient and the variables.

The coefficient of $a^{-6} b^4$ in the expansion is $\frac{1120}{27}$.

The final answer is $\boxed{\frac{1120}{27}}$.

Given:

The expansion is $(a^3 + ba)^{28}$.

To Find:

The middle term in the expansion.

Solution:

The given expansion is of the form $(x+y)^n$, where $x = a^3$, $y = ba$, and $n = 28$.

The total number of terms in the expansion is $n+1 = 28+1 = 29$.

Since the number of terms (29) is odd, there is one middle term.

The position of the middle term is $\frac{n}{2} + 1$ when $n$ is even.

Middle term position = $\frac{28}{2} + 1 = 14 + 1 = 15^{\text{th}}$ term.

We need to find the 15^th term ($T_{15}$) of the expansion.

The general term, $T_{k+1}$, in the expansion of $(x+y)^n$ is given by $T_{k+1} = \binom{n}{k} x^{n-k} y^k$.

For the 15^th term, $k+1=15$, so $k=14$.

Substitute the values of $n=28$, $k=14$, $x=a^3$, and $y=ba$ into the general term formula:

$T_{15} = \binom{28}{14} (a^3)^{28-14} (ba)^{14}$

$T_{15} = \binom{28}{14} (a^3)^{14} (ba)^{14}$

$T_{15} = \binom{28}{14} a^{3 \times 14} b^{14} a^{14}$

$T_{15} = \binom{28}{14} a^{42} b^{14} a^{14}$

Combine the terms involving $a$:

$T_{15} = \binom{28}{14} a^{42+14} b^{14}$

$T_{15} = \binom{28}{14} a^{56} b^{14}$

The middle term is $\binom{28}{14} a^{56} b^{14}$. We do not need to calculate the numerical value of $\binom{28}{14}$ unless specified, as it is part of the term's expression.

The middle term in the expansion of $(a^3 + ba)^{28}$ is $\binom{28}{14} a^{56} b^{14}$.

The final answer is $\boxed{\binom{28}{14} a^{56} b^{14}}$.

Given:

The expansion is $(1 + x)^{p + q}$.

$p$ and $q$ are positive integers (implied by the context of coefficients of $x^p$ and $x^q$ in this type of expansion). Let's assume they are non-negative integers as $x^0$ is possible.

To Find:

The ratio of the coefficient of $x^p$ to the coefficient of $x^q$ in the expansion.

Solution:

The expansion is of the form $(1+x)^n$, where $n = p+q$.

The coefficient of $x^k$ in the expansion of $(1+x)^n$ is $\binom{n}{k}$.

The coefficient of $x^p$ in the expansion of $(1 + x)^{p + q}$ is $\binom{p+q}{p}$.

The coefficient of $x^q$ in the expansion of $(1 + x)^{p + q}$ is $\binom{p+q}{q}$.

We need to find the ratio of these coefficients: $\frac{\text{Coefficient of } x^p}{\text{Coefficient of } x^q} = \frac{\binom{p+q}{p}}{\binom{p+q}{q}}$.

We know the symmetry property of binomial coefficients: $\binom{N}{K} = \binom{N}{N-K}$.

Here, $N=p+q$. Let's look at $\binom{p+q}{q}$ using this property:

$\binom{p+q}{q} = \binom{p+q}{(p+q) - q} = \binom{p+q}{p}$

So, the coefficient of $x^q$ is equal to the coefficient of $x^p$.

The ratio is $\frac{\binom{p+q}{p}}{\binom{p+q}{p}} = 1$, assuming $\binom{p+q}{p} \neq 0$. This is true for non-negative integers $p, q$ such that $p \le p+q$. If $p+q = 0$, the expansion is $(1+x)^0=1$. If $p=0, q=0$, the coefficient of $x^0$ is $\binom{0}{0}=1$. The ratio is $1/1=1$. If $p+q>0$, the coefficients are non-zero for $0 \le p, q \le p+q$.

The ratio of the coefficients is $1:1$, which is equal to 1.

The ratio of the coefficients of $x^p$ and $x^q$ in the expansion of $(1 + x)^{p + q}$ is 1.

The final answer is $\boxed{1}$.

Given:

The expansion is $\left( \sqrt{\frac{x}{3}} + \frac{3}{2x^2} \right)^{10}$.

To Find:

The position (term number) of the term independent of $x$.

Solution:

The given expansion is of the form $(a+b)^n$, where $a = \sqrt{\frac{x}{3}} = \frac{x^{1/2}}{\sqrt{3}} = \frac{1}{\sqrt{3}} x^{1/2}$, $b = \frac{3}{2x^2} = \frac{3}{2} x^{-2}$, and $n = 10$.

The general term, $T_{k+1}$, in the expansion of $(a+b)^n$ is given by the formula:

$T_{k+1} = \binom{n}{k} a^{n-k} b^k$

Substitute the values of $n=10$, $a=\frac{1}{\sqrt{3}} x^{1/2}$, and $b=\frac{3}{2} x^{-2}$ for the given expansion:

$T_{k+1} = \binom{10}{k} \left(\frac{1}{\sqrt{3}} x^{1/2}\right)^{10-k} \left(\frac{3}{2} x^{-2}\right)^k$

Simplify the terms involving $x$ and the constant terms separately.

Constant part: $\binom{10}{k} \left(\frac{1}{\sqrt{3}}\right)^{10-k} \left(\frac{3}{2}\right)^k$

Variable part: $(x^{1/2})^{10-k} \cdot (x^{-2})^k = x^{(1/2)(10-k)} \cdot x^{-2k} = x^{\frac{10-k}{2}} \cdot x^{-2k} = x^{\frac{10-k}{2} - 2k} = x^{\frac{10-k-4k}{2}} = x^{\frac{10-5k}{2}}$

The general term is $T_{k+1} = \binom{10}{k} \left(\frac{1}{\sqrt{3}}\right)^{10-k} \left(\frac{3}{2}\right)^k x^{\frac{10-5k}{2}}$.

For the term independent of $x$, the power of $x$ must be zero. So, we set the exponent of $x$ equal to 0:

$\frac{10-5k}{2} = 0$

$10 - 5k = 0$

$10 = 5k$

$k = 2$

The index $k$ determines the position of the term. The term number is $k+1$.

Term number = $k+1 = 2+1 = 3^{\text{rd}}$ term.

Since $k=2$ is a non-negative integer and $0 \leq 2 \leq 10$, this value of $k$ is valid, and a term independent of $x$ exists at this position.

The position of the term independent of $x$ is the 3^rd term.

The final answer is $\boxed{\text{3}^{\text{rd}} \text{ term}}$.

Given:

We need to find the remainder when $25^{15}$ is divided by 13.

To Find:

The remainder.

Solution:

We can use the concept of modular arithmetic or binomial expansion.

We want to find $25^{15} \pmod{13}$.

Notice that $25 = 2 \times 13 - 1 = 26 - 1$.

So, $25 \equiv -1 \pmod{13}$.

Now, raise both sides to the power of 15:

$25^{15} \equiv (-1)^{15} \pmod{13}$

Since 15 is an odd integer, $(-1)^{15} = -1$.

$25^{15} \equiv -1 \pmod{13}$

A remainder cannot be negative. To find the positive remainder, we add the modulus (13).

$-1 \equiv -1 + 13 \pmod{13}$

$-1 \equiv 12 \pmod{13}$

So, $25^{15} \equiv 12 \pmod{13}$.

The remainder when $25^{15}$ is divided by 13 is 12.

Alternate Solution (using Binomial Expansion):

We write 25 as a number related to 13. $25 = 26 - 1 = 2 \times 13 - 1$.

So, $25^{15} = (2 \times 13 - 1)^{15}$.

Let $x = 2 \times 13 = 26$ and $y = -1$. We expand $(x+y)^{15}$ using the binomial theorem:

$(26 - 1)^{15} = \binom{15}{0}(26)^{15}(-1)^0 + \binom{15}{1}(26)^{14}(-1)^1 + \binom{15}{2}(26)^{13}(-1)^2 + \dots + \binom{15}{14}(26)^1(-1)^{14} + \binom{15}{15}(26)^0(-1)^{15}$

$(26 - 1)^{15} = \binom{15}{0}26^{15} - \binom{15}{1}26^{14} + \binom{15}{2}26^{13} - \dots + \binom{15}{14}26 - \binom{15}{15}$

$(26 - 1)^{15} = \binom{15}{0}26^{15} - \binom{15}{1}26^{14} + \binom{15}{2}26^{13} - \dots + 15 \times 26 - 1$

Notice that every term in the expansion, except the last one $(-\binom{15}{15})$, has a factor of 26. Since $26$ is a multiple of 13, all these terms are divisible by 13.

$26^{15} - \binom{15}{1}26^{14} + \dots + \binom{15}{14}26 = \text{a multiple of } 26 = \text{a multiple of } 13$.

So, $(26 - 1)^{15} = (\text{a multiple of } 13) - \binom{15}{15}$

$(26 - 1)^{15} = (\text{a multiple of } 13) - 1$

Let $25^{15} = 13q + r$, where $q$ is the quotient and $r$ is the remainder, $0 \leq r < 13$.

$25^{15} = (\text{a multiple of } 13) - 1$

This means $25^{15} = 13k - 1$ for some integer $k$.

To find the remainder when divided by 13, we can write $13k - 1 = 13k - 13 + 12 = 13(k-1) + 12$.

So, $25^{15} = 13 \times (\text{an integer}) + 12$.

The remainder is 12.

When $25^{15}$ is divided by 13, the remainder is 12.

The final answer is $\boxed{12}$.

Given:

The statement: The sum of the series $\sum\limits^{10}_{r=0}\; \binom{20}{r}$ is $2^{19} + \frac{\binom{20}{10}}{2}$.

To Determine:

Whether the given statement is True or False.

Solution:

The given series is $S = \sum\limits^{10}_{r=0}\; \binom{20}{r} = \binom{20}{0} + \binom{20}{1} + \binom{20}{2} + \dots + \binom{20}{10}$.

We know the binomial theorem identity: $\sum\limits^{n}_{r=0}\; \binom{n}{r} = \binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n} = 2^n$.

For $n=20$, the total sum of coefficients is $\sum\limits^{20}_{r=0}\; \binom{20}{r} = 2^{20}$.

This sum can be split into two parts:

$\sum\limits^{20}_{r=0}\; \binom{20}{r} = \left( \binom{20}{0} + \binom{20}{1} + \dots + \binom{20}{9} \right) + \binom{20}{10} + \left( \binom{20}{11} + \binom{20}{12} + \dots + \binom{20}{20} \right)$

Using the symmetry property $\binom{n}{r} = \binom{n}{n-r}$:

$\binom{20}{0} = \binom{20}{20}$

$\binom{20}{1} = \binom{20}{19}$

... environments.

$\binom{20}{9} = \binom{20}{11}$

So, the terms $\binom{20}{0}, \dots, \binom{20}{9}$ are equal to $\binom{20}{20}, \dots, \binom{20}{11}$ respectively.

Let $S' = \binom{20}{0} + \binom{20}{1} + \dots + \binom{20}{9}$.

The sum $\sum\limits^{20}_{r=0}\; \binom{20}{r}$ can be written as:

$2^{20} = (\binom{20}{0} + \dots + \binom{20}{9}) + \binom{20}{10} + (\binom{20}{11} + \dots + \binom{20}{20})$

$2^{20} = S' + \binom{20}{10} + S'$ (since $\binom{20}{r} = \binom{20}{20-r}$)

$2^{20} = 2S' + \binom{20}{10}$

$2S' = 2^{20} - \binom{20}{10}$

$S' = \frac{1}{2} \left( 2^{20} - \binom{20}{10} \right) = 2^{19} - \frac{\binom{20}{10}}{2}$

The given series is $S = \sum\limits^{10}_{r=0}\; \binom{20}{r} = (\binom{20}{0} + \dots + \binom{20}{9}) + \binom{20}{10} = S' + \binom{20}{10}$.

Substitute the expression for $S'$:

$S = \left( 2^{19} - \frac{\binom{20}{10}}{2} \right) + \binom{20}{10}$

$S = 2^{19} + \binom{20}{10} - \frac{\binom{20}{10}}{2}$

$S = 2^{19} + \frac{\binom{20}{10}}{2}$

This matches the expression given in the statement.

The statement is True.

The final answer is $\boxed{True}$.

Given:

The statement: The expression $7^9 + 9^7$ is divisible by 64.

To Determine:

Whether the given statement is True or False.

Solution:

Let the expression be $E = 7^9 + 9^7$.

We can write $7 = 8 - 1$ and $9 = 8 + 1$.

$E = (8 - 1)^9 + (8 + 1)^7$

Let's expand $(x-y)^n$ and $(x+y)^n$ where $x=8$ and $y=1$.

$(x-y)^n = \binom{n}{0}x^n - \binom{n}{1}x^{n-1}y + \binom{n}{2}x^{n-2}y^2 - \dots + (-1)^n \binom{n}{n}y^n$

$(x+y)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}y + \binom{n}{2}x^{n-2}y^2 + \dots + \binom{n}{n}y^n$

Expand $7^9 = (8-1)^9$. Here $n=9, x=8, y=1$.

$7^9 = \binom{9}{0}8^9 - \binom{9}{1}8^8 + \binom{9}{2}8^7 - \binom{9}{3}8^6 + \binom{9}{4}8^5 - \binom{9}{5}8^4 + \binom{9}{6}8^3 - \binom{9}{7}8^2 + \binom{9}{8}8^1 - \binom{9}{9}8^0$

Expand $9^7 = (8+1)^7$. Here $n=7, x=8, y=1$.

$9^7 = \binom{7}{0}8^7 + \binom{7}{1}8^6 + \binom{7}{2}8^5 + \binom{7}{3}8^4 + \binom{7}{4}8^3 + \binom{7}{5}8^2 + \binom{7}{6}8^1 + \binom{7}{7}8^0$

Now, consider the sum $E = 7^9 + 9^7$. We need to check if it is divisible by 64. $64 = 8^2$.

Look at the terms in the expansions. Any term with $8^k$ where $k \ge 2$ is divisible by $8^2 = 64$.

In the expansion of $7^9$: All terms from $\binom{9}{0}8^9$ down to $\binom{9}{7}8^2$ are divisible by 64. The remaining terms are $\binom{9}{8}8^1 - \binom{9}{9}8^0$.

$7^9 = (\text{a multiple of } 64) + \binom{9}{8}8 - \binom{9}{9}$

$\binom{9}{8} = 9$, $\binom{9}{9} = 1$.

$7^9 = (\text{a multiple of } 64) + 9 \times 8 - 1 = (\text{a multiple of } 64) + 72 - 1 = (\text{a multiple of } 64) + 71$.

We can write $71 = 64 + 7$.

$7^9 = (\text{a multiple of } 64) + 64 + 7 = (\text{a multiple of } 64) + 7$.

So, $7^9 \equiv 7 \pmod{64}$.

In the expansion of $9^7$: All terms from $\binom{7}{0}8^7$ down to $\binom{7}{2}8^5$ are divisible by 64. (No, terms up to $\binom{7}{2}8^5$ are divisible by $8^5$. We need $8^2$. Terms with $8^k$ where $k \ge 2$ are divisible by 64). All terms from $\binom{7}{0}8^7$ down to $\binom{7}{2}8^5$ (incorrect indices). Terms are $\binom{7}{0}8^7, \binom{7}{1}8^6, \binom{7}{2}8^5, \dots$.

All terms with $8^k$ where $k \ge 2$ are divisible by 64. These are terms from $k=2$ up to $k=7$. The terms from $\binom{7}{0}8^7$ down to $\binom{7}{2}8^5$ are divisible by 64. The remaining terms are $\binom{7}{5}8^2 + \binom{7}{6}8^1 + \binom{7}{7}8^0$.

No, the powers of 8 are $7, 6, 5, 4, 3, 2, 1, 0$. Terms with $8^k$ for $k \ge 2$ are divisible by 64. These are terms with $k=0, 1, 2, 3, 4, 5$ in the binomial expansion of $(8+1)^7 = \sum \binom{7}{k} 8^{7-k} 1^k$. The power of 8 is $7-k$. We need $7-k \ge 2 \implies k \le 5$. So terms for $k=0, 1, 2, 3, 4, 5$ are divisible by 64.

The remaining terms are when $k=6$ and $k=7$.

$9^7 = (\text{a multiple of } 64) + \binom{7}{6}8^1 + \binom{7}{7}8^0$

$\binom{7}{6} = 7$, $\binom{7}{7} = 1$.

$9^7 = (\text{a multiple of } 64) + 7 \times 8 + 1 = (\text{a multiple of } 64) + 56 + 1 = (\text{a multiple of } 64) + 57$.

So, $9^7 \equiv 57 \pmod{64}$.

Now, sum the remainders:

$7^9 + 9^7 \equiv 7 + 57 \pmod{64}$

$7^9 + 9^7 \equiv 64 \pmod{64}$

$7^9 + 9^7 \equiv 0 \pmod{64}$

Since the remainder is 0, the expression $7^9 + 9^7$ is divisible by 64.

The statement is True.

The final answer is $\boxed{True}$.

Given:

The statement: The number of terms in the expansion of $[(2x + y^3)^4]^7$ is 8.

To Determine:

Whether the given statement is True or False.

Solution:

The given expression is $[(2x + y^3)^4]^7$.

Using the exponent rule $(a^m)^p = a^{mp}$, we can simplify the expression:

$[(2x + y^3)^4]^7 = (2x + y^3)^{4 \times 7} = (2x + y^3)^{28}$

This is a binomial expansion of the form $(A+B)^n$, where $A = 2x$, $B = y^3$, and $n = 28$.

The total number of terms in the expansion of $(A+B)^n$ is $n+1$.

In this case, the number of terms is $28 + 1 = 29$.

The statement claims that the number of terms is 8.

Since $29 \neq 8$, the statement is false.

The statement is False.

The final answer is $\boxed{False}$.

Given:

The statement: The sum of coefficients of the two middle terms in the expansion of $(1 + x)^{2n – 1}$ is equal to $\binom{2n – 1}{n}$.

To Determine:

Whether the given statement is True or False.

Solution:

The given expansion is $(1+x)^{2n-1}$. The power of the binomial is $N = 2n-1$.

The total number of terms in the expansion is $N+1 = (2n-1) + 1 = 2n$.

Since the number of terms ($2n$) is even (assuming $n \ge 1$), there are two middle terms.

The positions of the two middle terms are $\frac{2n}{2} = n^{\text{th}}$ term and $\frac{2n}{2} + 1 = (n+1)^{\text{th}}$ term.

The coefficient of the $m^{\text{th}}$ term in the expansion of $(1+x)^N$ is $\binom{N}{m-1}$.

The coefficient of the $n^{\text{th}}$ term is $\binom{2n-1}{n-1}$.

The coefficient of the $(n+1)^{\text{th}}$ term is $\binom{2n-1}{(n+1)-1} = \binom{2n-1}{n}$.

We need to find the sum of these two coefficients:

Sum of coefficients of middle terms = $\binom{2n-1}{n-1} + \binom{2n-1}{n}$.

Using Pascal's Identity: $\binom{N}{K} + \binom{N}{K+1} = \binom{N+1}{K+1}$.

Here, $N=2n-1$ and $K=n-1$. So $K+1 = n$.

$\binom{2n-1}{n-1} + \binom{2n-1}{n} = \binom{(2n-1)+1}{(n-1)+1} = \binom{2n}{n}$.

The sum of the coefficients of the two middle terms is $\binom{2n}{n}$.

The statement claims that the sum of coefficients of the two middle terms is equal to $\binom{2n-1}{n}$.

We found the sum is $\binom{2n}{n}$. We need to compare $\binom{2n}{n}$ with $\binom{2n-1}{n}$.

From Question 21, we found the ratio $\frac{\binom{2n}{n}}{\binom{2n-1}{n}} = 2$.

This means $\binom{2n}{n} = 2 \binom{2n-1}{n}$.

The sum of the coefficients of the middle terms is $2 \binom{2n-1}{n}$.

The statement claims the sum is $\binom{2n-1}{n}$. This is only true if $\binom{2n-1}{n} = 0$, which is not the case for $n \ge 1$.

For example, if $n=1$, the expansion is $(1+x)^{2(1)-1} = (1+x)^1 = 1+x$. The number of terms is 2. The middle terms are the 1st and 2nd terms. Coefficients are $\binom{1}{0}=1$ and $\binom{1}{1}=1$. Sum = $1+1=2$. The statement claims the sum is $\binom{2(1)-1}{1} = \binom{1}{1}=1$. $2 \neq 1$. Statement is false.

If $n=2$, the expansion is $(1+x)^{2(2)-1} = (1+x)^3 = 1+3x+3x^2+x^3$. The number of terms is 4. The middle terms are the 2nd and 3rd terms. Coefficients are $\binom{3}{1}=3$ and $\binom{3}{2}=3$. Sum = $3+3=6$. The statement claims the sum is $\binom{2(2)-1}{2} = \binom{3}{2}=3$. $6 \neq 3$. Statement is false.

The sum of the coefficients of the two middle terms is $\binom{2n}{n}$, which is equal to $2 \binom{2n-1}{n}$. The statement says the sum is $\binom{2n-1}{n}$.

The statement is False.

The statement is False.

The final answer is $\boxed{False}$.

Given:

The statement: The last two digits of the number $3^{400}$ are 01.

To Determine:

Whether the given statement is True or False.

Solution:

Finding the last two digits of a number is equivalent to finding the remainder when the number is divided by 100. So we need to calculate $3^{400} \pmod{100}$.

We can use Euler's totient theorem. $\phi(100) = \phi(2^2 \cdot 5^2) = \phi(2^2) \phi(5^2) = (2^2 - 2^1)(5^2 - 5^1) = (4-2)(25-5) = 2 \times 20 = 40$.

Since $\text{gcd}(3, 100) = 1$, by Euler's theorem, $3^{\phi(100)} \equiv 1 \pmod{100}$.

$3^{40} \equiv 1 \pmod{100}$.

We want to find $3^{400} \pmod{100}$.

$3^{400} = 3^{40 \times 10} = (3^{40})^{10}$.

Since $3^{40} \equiv 1 \pmod{100}$, we can raise both sides to the power of 10:

$(3^{40})^{10} \equiv 1^{10} \pmod{100}$

$3^{400} \equiv 1 \pmod{100}$.

This means the remainder when $3^{400}$ is divided by 100 is 1. The remainder 1 corresponds to the last two digits being 01.

The statement is True.

The final answer is $\boxed{True}$.

Given:

The statement: If the expansion of $\left( x - \frac{1}{x^2} \right)^{2n}$ contains a term independent of $x$, then $n$ is a multiple of 2.

To Determine:

Whether the given statement is True or False.

Solution:

The given expansion is of the form $(a+b)^N$, where $a = x$, $b = -\frac{1}{x^2} = -x^{-2}$, and $N = 2n$.

The general term, $T_{k+1}$, in the expansion of $(a+b)^N$ is given by the formula:

$T_{k+1} = \binom{N}{k} a^{N-k} b^k$

Substitute the values of $N=2n$, $a=x$, and $b=-x^{-2}$ for the given expansion:

$T_{k+1} = \binom{2n}{k} (x)^{2n-k} (-x^{-2})^k$

Simplify the terms involving $x$ and the constant terms separately.

Constant part: $\binom{2n}{k} (-1)^k$

Variable part: $(x)^{2n-k} \cdot (x^{-2})^k = x^{2n-k} \cdot x^{-2k} = x^{2n-k-2k} = x^{2n-3k}$

The general term is $T_{k+1} = \binom{2n}{k} (-1)^k x^{2n-3k}$.

For the term independent of $x$, the power of $x$ must be zero. So, we set the exponent of $x$ equal to 0:

$2n - 3k = 0$

$2n = 3k$

For a term independent of $x$ to exist, there must be a non-negative integer value of $k$ (where $0 \leq k \leq 2n$) that satisfies this equation.

From $2n = 3k$, we see that $3k$ must be an even number (since $2n$ is even). For $3k$ to be even, $k$ must be an even number (since 3 is odd). Let $k = 2m$ for some non-negative integer $m$.

Substitute $k=2m$ into the equation:

$2n = 3(2m)$

$2n = 6m$

$n = 3m$

So, if a term independent of $x$ exists, $n$ must be a multiple of 3 (i.e., $n$ can be written as $3m$ for some integer $m$).

Also, the value of $k = \frac{2n}{3}$ must be an integer such that $0 \leq k \leq 2n$. If $n = 3m$, then $k = \frac{2(3m)}{3} = 2m$. The condition $0 \leq k \leq 2n$ becomes $0 \leq 2m \leq 2(3m)$, which simplifies to $0 \leq 2m \leq 6m$. This is true for any non-negative integer $m$. So, a term independent of $x$ exists if and only if $n$ is a multiple of 3.

The statement says that if a term independent of $x$ exists, then $n$ is a multiple of 2.

We found that $n$ must be a multiple of 3.

Let's check a case where $n$ is a multiple of 3 but not a multiple of 2. Let $n=3$. The expansion is $\left( x - \frac{1}{x^2} \right)^{2 \times 3} = \left( x - \frac{1}{x^2} \right)^6$. The term independent of $x$ exists because $n=3$ is a multiple of 3. The corresponding $k = \frac{2n}{3} = \frac{2(3)}{3} = 2$. $T_3 = \binom{6}{2} (x)^4 (-x^{-2})^2 = 15 x^4 x^{-4} = 15$. A term independent of $x$ exists.

According to the statement, since a term independent of $x$ exists, $n$ must be a multiple of 2. But $n=3$ is not a multiple of 2. This contradicts the statement.

The statement is False.

The statement is False.

The final answer is $\boxed{False}$.

Given:

The statement: The number of terms in the expansion of $(a + b)^n$ where $n \in \mathbb{N}$ is one less than the power $n$.

To Determine:

Whether the given statement is True or False.

Solution:

The binomial theorem states that the expansion of $(a+b)^n$ is given by:

$(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1} b^1 + \binom{n}{2}a^{n-2} b^2 + \dots + \binom{n}{n}a^0 b^n$

The terms in the expansion correspond to the values of the index in the binomial coefficient $k = 0, 1, 2, \dots, n$.

The number of possible values for $k$ is the number of integers from 0 to $n$, inclusive. This is $n - 0 + 1 = n+1$.

So, the total number of terms in the expansion of $(a+b)^n$ is $n+1$.

The statement claims that the number of terms is "one less than the power $n$", which means $n-1$.

We found the number of terms is $n+1$.

Since $n+1 \neq n-1$ for any $n \in \mathbb{N}$, the statement is false.

For example, if $n=1$, $(a+b)^1 = a+b$, which has 2 terms. The power is 1. $1+1 = 2$. $1-1=0$. The statement claims 0 terms.

If $n=2$, $(a+b)^2 = a^2 + 2ab + b^2$, which has 3 terms. The power is 2. $2+1 = 3$. $2-1=1$. The statement claims 1 term.

The statement is False.

The final answer is $\boxed{False}$.