Chapter 7 Binomial Theorem (Class 11 - Latest Maths NCERT Textbook Solutions)

Solution:

We need to expand the expression $\left( x^{2} +\frac{3}{x}\right)^{4}$. This is in the form $(a+b)^n$, where $a = x^2$, $b = \frac{3}{x}$, and $n=4$.

We can use the Binomial Theorem for expansion. The Binomial Theorem states that for any positive integer $n$, the expansion of $(a+b)^n$ is given by:

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

where $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ are the binomial coefficients.

For the given expression, $a = x^2$, $b = \frac{3}{x}$, and $n = 4$. We need to calculate the terms for $r=0, 1, 2, 3, 4$ and sum them up.

Let's calculate each term:

Term for $r=0$:

$\binom{4}{0}(x^2)^{4-0}\left(\frac{3}{x}\right)^0 = 1 \cdot (x^2)^4 \cdot 1 = x^8$

Term for $r=1$:

$\binom{4}{1}(x^2)^{4-1}\left(\frac{3}{x}\right)^1 = 4 \cdot (x^2)^3 \cdot \frac{3}{x} = 4 \cdot x^6 \cdot \frac{3}{x} = 12x^{6-1} = 12x^5$

Term for $r=2$:

$\binom{4}{2}(x^2)^{4-2}\left(\frac{3}{x}\right)^2 = 6 \cdot (x^2)^2 \cdot \left(\frac{3^2}{x^2}\right) = 6 \cdot x^4 \cdot \frac{9}{x^2} = 54x^{4-2} = 54x^2$

Term for $r=3$:

$\binom{4}{3}(x^2)^{4-3}\left(\frac{3}{x}\right)^3 = 4 \cdot (x^2)^1 \cdot \left(\frac{3^3}{x^3}\right) = 4 \cdot x^2 \cdot \frac{27}{x^3} = 108x^{2-3} = 108x^{-1} = \frac{108}{x}$

Term for $r=4$:

$\binom{4}{4}(x^2)^{4-4}\left(\frac{3}{x}\right)^4 = 1 \cdot (x^2)^0 \cdot \left(\frac{3^4}{x^4}\right) = 1 \cdot 1 \cdot \frac{81}{x^4} = \frac{81}{x^4}$

Now, we sum all the terms to get the complete expansion:

$\left( x^{2} +\frac{3}{x}\right)^{4} = (\text{Term for } r=0) + (\text{Term for } r=1) + (\text{Term for } r=2) + (\text{Term for } r=3) + (\text{Term for } r=4)$

$\left( x^{2} +\frac{3}{x}\right)^{4} = x^8 + 12x^5 + 54x^2 + \frac{108}{x} + \frac{81}{x^4}$

The final expansion is:

$\left( x^{2} +\frac{3}{x}\right)^{4} = x^8 + 12x^5 + 54x^2 + \frac{108}{x} + \frac{81}{x^4}$

Given:

We need to compute the value of $(98)^5$.

To Compute:

$(98)^5$

Solution:

We can rewrite $98$ as $100 - 2$. So, $(98)^5$ can be written as $(100 - 2)^5$.

This expression is in the form $(a-b)^n$, where $a = 100$, $b = 2$, and $n = 5$. We can use the Binomial Theorem to expand this expression.

The Binomial Theorem for $(a-b)^n$ is given by:

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

Using this formula for $(100 - 2)^5$ ($n=5$, $a=100$, $b=2$):

$(100 - 2)^5 = \binom{5}{0}(100)^5(-2)^0 + \binom{5}{1}(100)^4(-2)^1 + \binom{5}{2}(100)^3(-2)^2 + \binom{5}{3}(100)^2(-2)^3 + \binom{5}{4}(100)^1(-2)^4 + \binom{5}{5}(100)^0(-2)^5$

Let's calculate each term:

Term 1 (r=0):

$\binom{5}{0}(100)^5(-2)^0 = 1 \times (10000000000) \times 1 = 10,00,00,00,000$

Term 2 (r=1):

$\binom{5}{1}(100)^4(-2)^1 = 5 \times (100000000) \times (-2) = 5 \times (-200000000) = -1,00,00,00,000$

Term 3 (r=2):

$\binom{5}{2}(100)^3(-2)^2 = 10 \times (1000000) \times 4 = 10 \times (4000000) = 4,00,00,000$

Term 4 (r=3):

$\binom{5}{3}(100)^2(-2)^3 = 10 \times (10000) \times (-8) = 10 \times (-80000) = -8,00,000$

Term 5 (r=4):

$\binom{5}{4}(100)^1(-2)^4 = 5 \times (100) \times 16 = 5 \times 1600 = 8,000$

Term 6 (r=5):

$\binom{5}{5}(100)^0(-2)^5 = 1 \times 1 \times (-32) = -32$

Now, we add all the terms together:

$(98)^5 = 10,00,00,00,000 + (-1,00,00,00,000) + 4,00,00,000 + (-8,00,000) + 8,000 + (-32)$

$(98)^5 = 10,00,00,00,000 - 1,00,00,00,000 + 4,00,00,000 - 8,00,000 + 8,000 - 32$

$(98)^5 = 9,00,00,00,000 + 4,00,00,000 - 8,00,000 + 8,000 - 32$

$(98)^5 = 9,04,00,00,000 - 8,00,000 + 8,000 - 32$

$(98)^5 = 9,03,92,00,000 + 8,000 - 32$

$(98)^5 = 9,03,92,08,000 - 32$

$(98)^5 = 9,03,92,07,968$

The value of $(98)^5$ is $9,03,92,07,968$.

Solution:

We are asked to compare $(1.01)^{1000000}$ and $10,000$.

Let's consider the expression $(1.01)^{1000000}$. We can write $1.01$ as $1 + 0.01$. So, we need to evaluate $(1 + 0.01)^{1000000}$.

This is in the form $(1+x)^n$, where $x = 0.01$ and $n = 1000000$. We can use the Binomial Theorem to expand this expression.

The Binomial Theorem for $(1+x)^n$, where $n$ is a positive integer, is given by:

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

Substituting $n = 1000000$ and $x = 0.01$, we get:

$(1 + 0.01)^{1000000} = \binom{1000000}{0} + \binom{1000000}{1}(0.01) + \binom{1000000}{2}(0.01)^2 + \binom{1000000}{3}(0.01)^3 + \dots$

Let's calculate the first few terms of the expansion:

The first term is $\binom{1000000}{0} = 1$.

The second term is $\binom{1000000}{1}(0.01) = 1000000 \cdot (0.01) = 1000000 \cdot \frac{1}{100} = 10000$.

The third term is $\binom{1000000}{2}(0.01)^2 = \frac{1000000 \cdot (1000000-1)}{2 \cdot 1} (0.0001) = \frac{1000000 \cdot 999999}{2} \cdot \frac{1}{10000}$.

Simplifying the third term: $\frac{1000000}{10000} = 100$. So the third term is $\frac{100 \cdot 999999}{2} = 50 \cdot 999999 = 49999950$.

So, the expansion begins as:

$(1.01)^{1000000} = 1 + 10000 + 49999950 + (\text{subsequent terms})$

Since $n=1000000$ is a positive integer and $x=0.01$ is positive, all the terms $\binom{n}{r}x^r$ in the expansion are positive.

Therefore, $(1.01)^{1000000} = 1 + 10000 + (\text{sum of positive terms})$.

This means $(1.01)^{1000000} = 10001 + (\text{sum of positive terms})$.

Since the sum of the remaining terms is positive, we have:

$(1.01)^{1000000} > 10001$

We are comparing $(1.01)^{1000000}$ with $10000$.

Since $10001 > 10000$, and $(1.01)^{1000000} > 10001$, by transitivity, we can conclude that $(1.01)^{1000000}$ is larger than $10000$.

Thus, $(1.01)^{1000000} > 10000$.

Conclusion: $(1.01)^{1000000}$ is larger than $10,000$.

Given:

The expression $6^n - 5n$, where $n$ is a non-negative integer.

To Prove:

That $6^n - 5n$ always leaves a remainder of $1$ when divided by $25$.

This means we need to show that $6^n - 5n = 25k + 1$ for some integer $k$, for all non-negative integers $n$.

Proof:

We will use the Binomial Theorem to expand $6^n$. We can write $6$ as $1+5$.

Using the Binomial Theorem for $(a+b)^n$, with $a=1$ and $b=5$:

Simplify the terms, knowing that $1^k = 1$ for any $k$ and $5^0=1$:

$(1+5)^n = \binom{n}{0} + \binom{n}{1}5 + \binom{n}{2}5^2 + \binom{n}{3}5^3 + \dots + \binom{n}{n}5^n$

We know that $\binom{n}{0} = 1$ and $\binom{n}{1} = n$. Substitute these values:

Now, consider the expression $6^n - 5n$. Subtract $5n$ from both sides of equation (i):

Consider the sum of terms $\sum\limits_{r=2}^{n} \binom{n}{r}5^r = \binom{n}{2}5^2 + \binom{n}{3}5^3 + \dots + \binom{n}{n}5^n$. We need to show that this sum is always divisible by $25$ for any non-negative integer $n$.

Case 1: $n = 0$

From equation (ii), $6^0 - 5(0) = 1 + \sum\limits_{r=2}^{0} \binom{0}{r}5^r$. The sum $\sum\limits_{r=2}^{0}$ is an empty sum, which is equal to $0$.

$6^0 - 5(0) = 1 + 0 = 1$.

$1$ divided by $25$ leaves a remainder of $1$.

Case 2: $n = 1$

From equation (ii), $6^1 - 5(1) = 1 + \sum\limits_{r=2}^{1} \binom{1}{r}5^r$. The sum $\sum\limits_{r=2}^{1}$ is an empty sum, which is equal to $0$.

$6^1 - 5(1) = 1 + 0 = 1$.

$1$ divided by $25$ leaves a remainder of $1$.

Case 3: $n \ge 2$

In this case, the sum $\sum\limits_{r=2}^{n} \binom{n}{r}5^r$ has terms.

The sum is $\binom{n}{2}5^2 + \binom{n}{3}5^3 + \dots + \binom{n}{n}5^n$.

Every term $\binom{n}{r}5^r$ in this sum (for $r \ge 2$) has a factor of $5^r$. Since $r \ge 2$, $5^r$ is divisible by $5^2 = 25$.

So, we can write each term as $\binom{n}{r}5^r = \binom{n}{r} 5^{r-2} \cdot 5^2 = 25 \cdot \binom{n}{r} 5^{r-2}$.

Therefore, the sum is:

$\sum\limits_{r=2}^{n} \binom{n}{r}5^r = 25 \cdot \binom{n}{2}5^0 + 25 \cdot \binom{n}{3}5^1 + \dots + 25 \cdot \binom{n}{n}5^{n-2}$

Factor out $25$:

Let $K = \binom{n}{2} + \binom{n}{3}5 + \dots + \binom{n}{n}5^{n-2} = \sum\limits_{r=2}^{n} \binom{n}{r}5^{r-2}$.

For $n \ge 2$, the binomial coefficients $\binom{n}{r}$ are integers for $r \ge 2$. The powers of $5$ ($5^{r-2}$ for $r \ge 2$, i.e., $5^0, 5^1, \dots$) are also integers.

The sum and product of integers are integers. Therefore, $K$ is an integer for $n \ge 2$.

From (iii), $\sum\limits_{r=2}^{n} \binom{n}{r}5^r = 25K$, where $K$ is an integer.

Substitute this back into equation (ii):

This shows that for $n \ge 2$, $6^n - 5n$ is in the form $25K + 1$, meaning it leaves a remainder of $1$ when divided by $25$.

Combining all cases ($n=0, n=1, n \ge 2$), we see that $6^n - 5n$ can always be written in the form $25k + 1$ for some integer $k$ (where $k=0$ for $n=0, 1$ and $k=K$ for $n \ge 2$).

Conclusion:

In all cases, when $6^n - 5n$ is divided by $25$, the remainder is $1$. Thus, $6^n - 5n$ always leaves remainder $1$ when divided by $25$.

Solution:

We need to expand the expression $(1 - 2x)^5$. This is in the form of $(a+b)^n$, where $a=1$, $b=-2x$, and $n=5$.

We can use the Binomial Theorem, which states that for a positive integer $n$,

$(a+b)^n = \sum\limits_{r=0}^{n} \binom{n}{r} a^{n-r} b^{r}$

Substituting $a=1$, $b=-2x$, and $n=5$, we get:

$(1 - 2x)^5 = \sum\limits_{r=0}^{5} \binom{5}{r} (1)^{5-r} (-2x)^{r}$

$(1 - 2x)^5 = \binom{5}{0}(1)^5(-2x)^0 + \binom{5}{1}(1)^4(-2x)^1 + \binom{5}{2}(1)^3(-2x)^2 + \binom{5}{3}(1)^2(-2x)^3 + \binom{5}{4}(1)^1(-2x)^4 + \binom{5}{5}(1)^0(-2x)^5$

Now, we calculate each term:

Term 1 (r=0): $\binom{5}{0}(1)^5(-2x)^0 = 1 \cdot 1 \cdot 1 = 1$

Term 2 (r=1): $\binom{5}{1}(1)^4(-2x)^1 = 5 \cdot 1 \cdot (-2x) = -10x$

Term 3 (r=2): $\binom{5}{2}(1)^3(-2x)^2 = 10 \cdot 1 \cdot (4x^2) = 40x^2$

Term 4 (r=3): $\binom{5}{3}(1)^2(-2x)^3 = 10 \cdot 1 \cdot (-8x^3) = -80x^3$

Term 5 (r=4): $\binom{5}{4}(1)^1(-2x)^4 = 5 \cdot 1 \cdot (16x^4) = 80x^4$

Term 6 (r=5): $\binom{5}{5}(1)^0(-2x)^5 = 1 \cdot 1 \cdot (-32x^5) = -32x^5$

Summing these terms, we get the expansion:

$(1 - 2x)^5 = 1 + (-10x) + 40x^2 + (-80x^3) + 80x^4 + (-32x^5)$

$(1 - 2x)^5 = 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5$

The expanded form of $(1 - 2x)^5$ is $1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5$.

Solution:

We need to expand the expression $\left( \frac{2}{x}-\frac{x}{2} \right)^{5}$. This is in the form of $(a+b)^n$, where $a=\frac{2}{x}$, $b=-\frac{x}{2}$, and $n=5$.

We use the Binomial Theorem, which states that:

$(a+b)^n = \sum\limits_{r=0}^{n} \binom{n}{r} a^{n-r} b^{r}$

Substituting $a=\frac{2}{x}$, $b=-\frac{x}{2}$, and $n=5$, we get:

$\left( \frac{2}{x}-\frac{x}{2} \right)^{5} = \sum\limits_{r=0}^{5} \binom{5}{r} \left(\frac{2}{x}\right)^{5-r} \left(-\frac{x}{2}\right)^{r}$

This expands to:

$\binom{5}{0}\left(\frac{2}{x}\right)^5\left(-\frac{x}{2}\right)^0 + \binom{5}{1}\left(\frac{2}{x}\right)^4\left(-\frac{x}{2}\right)^1 + \binom{5}{2}\left(\frac{2}{x}\right)^3\left(-\frac{x}{2}\right)^2 + \binom{5}{3}\left(\frac{2}{x}\right)^2\left(-\frac{x}{2}\right)^3 + \binom{5}{4}\left(\frac{2}{x}\right)^1\left(-\frac{x}{2}\right)^4 + \binom{5}{5}\left(\frac{2}{x}\right)^0\left(-\frac{x}{2}\right)^5$

Now, we calculate each term:

Term for $r=0$: $\binom{5}{0}\left(\frac{2}{x}\right)^5\left(-\frac{x}{2}\right)^0 = 1 \cdot \frac{2^5}{x^5} \cdot 1 = \frac{32}{x^5}$

Term for $r=1$: $\binom{5}{1}\left(\frac{2}{x}\right)^4\left(-\frac{x}{2}\right)^1 = 5 \cdot \frac{2^4}{x^4} \cdot \left(-\frac{x}{2}\right) = 5 \cdot \frac{16}{x^4} \cdot \left(-\frac{x}{2}\right) = -\frac{5 \cdot 16 \cdot x}{2 \cdot x^4} = -\frac{40}{x^3}$

Term for $r=2$: $\binom{5}{2}\left(\frac{2}{x}\right)^3\left(-\frac{x}{2}\right)^2 = 10 \cdot \frac{2^3}{x^3} \cdot \frac{x^2}{2^2} = 10 \cdot \frac{8}{x^3} \cdot \frac{x^2}{4} = \frac{10 \cdot 8 \cdot x^2}{4 \cdot x^3} = \frac{20}{x}$

Term for $r=3$: $\binom{5}{3}\left(\frac{2}{x}\right)^2\left(-\frac{x}{2}\right)^3 = 10 \cdot \frac{2^2}{x^2} \cdot \left(-\frac{x^3}{2^3}\right) = 10 \cdot \frac{4}{x^2} \cdot \left(-\frac{x^3}{8}\right) = -\frac{10 \cdot 4 \cdot x^3}{8 \cdot x^2} = -5x$

Term for $r=4$: $\binom{5}{4}\left(\frac{2}{x}\right)^1\left(-\frac{x}{2}\right)^4 = 5 \cdot \frac{2}{x} \cdot \frac{x^4}{2^4} = 5 \cdot \frac{2}{x} \cdot \frac{x^4}{16} = \frac{5 \cdot 2 \cdot x^4}{16 \cdot x} = \frac{10x^3}{16} = \frac{5x^3}{8}$

Term for $r=5$: $\binom{5}{5}\left(\frac{2}{x}\right)^0\left(-\frac{x}{2}\right)^5 = 1 \cdot 1 \cdot \left(-\frac{x^5}{2^5}\right) = -\frac{x^5}{32}$

Summing these terms, we get the expansion:

$\left( \frac{2}{x}-\frac{x}{2} \right)^{5} = \frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5x^3}{8} - \frac{x^5}{32}$

The expanded form of $\left( \frac{2}{x}-\frac{x}{2} \right)^{5}$ is $\frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5x^3}{8} - \frac{x^5}{32}$.

Solution:

We need to expand the expression $(2x - 3)^6$. This is in the form of $(a+b)^n$, where $a=2x$, $b=-3$, and $n=6$.

We use the Binomial Theorem, which states that for a positive integer $n$,

$(a+b)^n = \sum\limits_{r=0}^{n} \binom{n}{r} a^{n-r} b^{r}$

Substituting $a=2x$, $b=-3$, and $n=6$, we get:

$(2x - 3)^6 = \sum\limits_{r=0}^{6} \binom{6}{r} (2x)^{6-r} (-3)^{r}$

This expands to:

$\binom{6}{0}(2x)^6(-3)^0 + \binom{6}{1}(2x)^5(-3)^1 + \binom{6}{2}(2x)^4(-3)^2 + \binom{6}{3}(2x)^3(-3)^3 + \binom{6}{4}(2x)^2(-3)^4 + \binom{6}{5}(2x)^1(-3)^5 + \binom{6}{6}(2x)^0(-3)^6$

Now, we calculate each term:

Term for $r=0$: $\binom{6}{0}(2x)^6(-3)^0 = 1 \cdot (2^6 x^6) \cdot 1 = 1 \cdot 64x^6 \cdot 1 = 64x^6$

Term for $r=1$: $\binom{6}{1}(2x)^5(-3)^1 = 6 \cdot (2^5 x^5) \cdot (-3) = 6 \cdot 32x^5 \cdot (-3) = -576x^5$

Term for $r=2$: $\binom{6}{2}(2x)^4(-3)^2 = 15 \cdot (2^4 x^4) \cdot 9 = 15 \cdot 16x^4 \cdot 9 = 2160x^4$

Term for $r=3$: $\binom{6}{3}(2x)^3(-3)^3 = 20 \cdot (2^3 x^3) \cdot (-27) = 20 \cdot 8x^3 \cdot (-27) = -4320x^3$

Term for $r=4$: $\binom{6}{4}(2x)^2(-3)^4 = 15 \cdot (2^2 x^2) \cdot 81 = 15 \cdot 4x^2 \cdot 81 = 4860x^2$

Term for $r=5$: $\binom{6}{5}(2x)^1(-3)^5 = 6 \cdot (2x) \cdot (-243) = 12x \cdot (-243) = -2916x$

Term for $r=6$: $\binom{6}{6}(2x)^0(-3)^6 = 1 \cdot 1 \cdot 729 = 729$

Summing these terms, we get the expansion:

$(2x - 3)^6 = 64x^6 + (-576x^5) + 2160x^4 + (-4320x^3) + 4860x^2 + (-2916x) + 729$

$(2x - 3)^6 = 64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729$

The expanded form of $(2x - 3)^6$ is $64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729$.

Solution:

We need to expand the expression $\left( \frac{x}{3}+\frac{1}{x} \right)^{5}$. This is in the form of $(a+b)^n$, where $a=\frac{x}{3}$, $b=\frac{1}{x}$, and $n=5$.

We use the Binomial Theorem, which states that for a positive integer $n$,

$(a+b)^n = \sum\limits_{r=0}^{n} \binom{n}{r} a^{n-r} b^{r}$

Substituting $a=\frac{x}{3}$, $b=\frac{1}{x}$, and $n=5$, we get:

$\left( \frac{x}{3}+\frac{1}{x} \right)^{5} = \sum\limits_{r=0}^{5} \binom{5}{r} \left(\frac{x}{3}\right)^{5-r} \left(\frac{1}{x}\right)^{r}$

This expands to:

$\binom{5}{0}\left(\frac{x}{3}\right)^5\left(\frac{1}{x}\right)^0 + \binom{5}{1}\left(\frac{x}{3}\right)^4\left(\frac{1}{x}\right)^1 + \binom{5}{2}\left(\frac{x}{3}\right)^3\left(\frac{1}{x}\right)^2 + \binom{5}{3}\left(\frac{x}{3}\right)^2\left(\frac{1}{x}\right)^3 + \binom{5}{4}\left(\frac{x}{3}\right)^1\left(\frac{1}{x}\right)^4 + \binom{5}{5}\left(\frac{x}{3}\right)^0\left(\frac{1}{x}\right)^5$

Now, we calculate each term:

Term for $r=0$: $\binom{5}{0}\left(\frac{x}{3}\right)^5\left(\frac{1}{x}\right)^0 = 1 \cdot \frac{x^5}{3^5} \cdot 1 = \frac{x^5}{243}$

Term for $r=1$: $\binom{5}{1}\left(\frac{x}{3}\right)^4\left(\frac{1}{x}\right)^1 = 5 \cdot \frac{x^4}{3^4} \cdot \frac{1}{x} = 5 \cdot \frac{x^4}{81} \cdot \frac{1}{x} = \frac{5x^{4-1}}{81} = \frac{5x^3}{81}$

Term for $r=2$: $\binom{5}{2}\left(\frac{x}{3}\right)^3\left(\frac{1}{x}\right)^2 = 10 \cdot \frac{x^3}{3^3} \cdot \frac{1}{x^2} = 10 \cdot \frac{x^3}{27} \cdot \frac{1}{x^2} = \frac{10x^{3-2}}{27} = \frac{10x}{27}$

Term for $r=3$: $\binom{5}{3}\left(\frac{x}{3}\right)^2\left(\frac{1}{x}\right)^3 = 10 \cdot \frac{x^2}{3^2} \cdot \frac{1}{x^3} = 10 \cdot \frac{x^2}{9} \cdot \frac{1}{x^3} = \frac{10x^{2-3}}{9} = \frac{10x^{-1}}{9} = \frac{10}{9x}$

Term for $r=4$: $\binom{5}{4}\left(\frac{x}{3}\right)^1\left(\frac{1}{x}\right)^4 = 5 \cdot \frac{x}{3} \cdot \frac{1}{x^4} = \frac{5x^{1-4}}{3} = \frac{5x^{-3}}{3} = \frac{5}{3x^3}$

Term for $r=5$: $\binom{5}{5}\left(\frac{x}{3}\right)^0\left(\frac{1}{x}\right)^5 = 1 \cdot 1 \cdot \frac{1}{x^5} = \frac{1}{x^5}$

Summing these terms, we get the expansion:

$\left( \frac{x}{3}+\frac{1}{x} \right)^{5} = \frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}$

The expanded form of $\left( \frac{x}{3}+\frac{1}{x} \right)^{5}$ is $\frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}$.

Solution:

We need to expand the expression $\left( x+\frac{1}{x} \right)^{6}$. This is in the form of $(a+b)^n$, where $a=x$, $b=\frac{1}{x}$, and $n=6$.

We use the Binomial Theorem, which states that for a positive integer $n$,

$(a+b)^n = \sum\limits_{r=0}^{n} \binom{n}{r} a^{n-r} b^{r}$

Substituting $a=x$, $b=\frac{1}{x}$, and $n=6$, we get:

$\left( x+\frac{1}{x} \right)^{6} = \sum\limits_{r=0}^{6} \binom{6}{r} (x)^{6-r} \left(\frac{1}{x}\right)^{r}$

This expands to:

$\binom{6}{0}x^6\left(\frac{1}{x}\right)^0 + \binom{6}{1}x^5\left(\frac{1}{x}\right)^1 + \binom{6}{2}x^4\left(\frac{1}{x}\right)^2 + \binom{6}{3}x^3\left(\frac{1}{x}\right)^3 + \binom{6}{4}x^2\left(\frac{1}{x}\right)^4 + \binom{6}{5}x^1\left(\frac{1}{x}\right)^5 + \binom{6}{6}x^0\left(\frac{1}{x}\right)^6$

Now, we calculate each term using the binomial coefficients: $\binom{6}{0}=1, \binom{6}{1}=6, \binom{6}{2}=15, \binom{6}{3}=20, \binom{6}{4}=15, \binom{6}{5}=6, \binom{6}{6}=1$.

Term for $r=0$: $\binom{6}{0}x^6\left(\frac{1}{x}\right)^0 = 1 \cdot x^6 \cdot 1 = x^6$

Term for $r=1$: $\binom{6}{1}x^5\left(\frac{1}{x}\right)^1 = 6 \cdot x^5 \cdot x^{-1} = 6x^{5-1} = 6x^4$

Term for $r=2$: $\binom{6}{2}x^4\left(\frac{1}{x}\right)^2 = 15 \cdot x^4 \cdot x^{-2} = 15x^{4-2} = 15x^2$

Term for $r=3$: $\binom{6}{3}x^3\left(\frac{1}{x}\right)^3 = 20 \cdot x^3 \cdot x^{-3} = 20x^{3-3} = 20x^0 = 20$

Term for $r=4$: $\binom{6}{4}x^2\left(\frac{1}{x}\right)^4 = 15 \cdot x^2 \cdot x^{-4} = 15x^{2-4} = 15x^{-2} = \frac{15}{x^2}$

Term for $r=5$: $\binom{6}{5}x^1\left(\frac{1}{x}\right)^5 = 6 \cdot x^1 \cdot x^{-5} = 6x^{1-5} = 6x^{-4} = \frac{6}{x^4}$

Term for $r=6$: $\binom{6}{6}x^0\left(\frac{1}{x}\right)^6 = 1 \cdot x^0 \cdot x^{-6} = 1 \cdot 1 \cdot x^{-6} = \frac{1}{x^6}$

Summing these terms, we get the expansion:

$\left( x+\frac{1}{x} \right)^{6} = x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}$

The expanded form of $\left( x+\frac{1}{x} \right)^{6}$ is $x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}$.

Solution:

We need to evaluate $(96)^3$ using the binomial theorem.

We can write 96 as $100 - 4$. So, $(96)^3 = (100 - 4)^3$.

This is in the form of $(a-b)^n$, where $a=100$, $b=4$, and $n=3$.

We use the Binomial Theorem for $(a-b)^n$, which states:

$(a-b)^n = \sum\limits_{r=0}^{n} \binom{n}{r} a^{n-r} (-b)^{r}$

For $n=3$, the expansion is:

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

Substituting $a=100$ and $b=4$, we get:

$(100-4)^3 = \binom{3}{0}(100)^3(-4)^0 + \binom{3}{1}(100)^2(-4)^1 + \binom{3}{2}(100)^1(-4)^2 + \binom{3}{3}(100)^0(-4)^3$

Now, we calculate each term using the binomial coefficients: $\binom{3}{0}=1, \binom{3}{1}=3, \binom{3}{2}=3, \binom{3}{3}=1$.

Term 1 (r=0): $\binom{3}{0}(100)^3(-4)^0 = 1 \cdot 1,000,000 \cdot 1 = 1,000,000$

Term 2 (r=1): $\binom{3}{1}(100)^2(-4)^1 = 3 \cdot 10,000 \cdot (-4) = -120,000$

Term 3 (r=2): $\binom{3}{2}(100)^1(-4)^2 = 3 \cdot 100 \cdot 16 = 300 \cdot 16 = 4800$

Term 4 (r=3): $\binom{3}{3}(100)^0(-4)^3 = 1 \cdot 1 \cdot (-64) = -64$

Summing these terms, we get the value of $(96)^3$:

$(96)^3 = 1,000,000 + (-120,000) + 4800 + (-64)$

$(96)^3 = 1,000,000 - 120,000 + 4800 - 64$

$(96)^3 = 880,000 + 4800 - 64$

$(96)^3 = 884,800 - 64$

$(96)^3 = 884,736$

The value of $(96)^3$ evaluated using the binomial theorem is $884,736$.

Solution:

We need to evaluate $(102)^5$ using the binomial theorem.

We can write 102 as $100 + 2$. So, $(102)^5 = (100 + 2)^5$.

This is in the form of $(a+b)^n$, where $a=100$, $b=2$, and $n=5$.

We use the Binomial Theorem, which states that for a positive integer $n$,

$(a+b)^n = \sum\limits_{r=0}^{n} \binom{n}{r} a^{n-r} b^{r}$

For $n=5$, the expansion is:

$(100+2)^5 = \binom{5}{0}(100)^5(2)^0 + \binom{5}{1}(100)^4(2)^1 + \binom{5}{2}(100)^3(2)^2 + \binom{5}{3}(100)^2(2)^3 + \binom{5}{4}(100)^1(2)^4 + \binom{5}{5}(100)^0(2)^5$

Now, we calculate each term using the binomial coefficients: $\binom{5}{0}=1, \binom{5}{1}=5, \binom{5}{2}=10, \binom{5}{3}=10, \binom{5}{4}=5, \binom{5}{5}=1$.

Term for $r=0$: $\binom{5}{0}(100)^5(2)^0 = 1 \cdot (10000000000) \cdot 1 = 10,000,000,000$

Term for $r=1$: $\binom{5}{1}(100)^4(2)^1 = 5 \cdot (100000000) \cdot 2 = 10 \cdot 100,000,000 = 1,000,000,000$

Term for $r=2$: $\binom{5}{2}(100)^3(2)^2 = 10 \cdot (1000000) \cdot 4 = 40 \cdot 1,000,000 = 40,000,000$

Term for $r=3$: $\binom{5}{3}(100)^2(2)^3 = 10 \cdot (10000) \cdot 8 = 80 \cdot 10,000 = 800,000$

Term for $r=4$: $\binom{5}{4}(100)^1(2)^4 = 5 \cdot (100) \cdot 16 = 500 \cdot 16 = 8000$

Term for $r=5$: $\binom{5}{5}(100)^0(2)^5 = 1 \cdot 1 \cdot 32 = 32$

Summing these terms, we get the value of $(102)^5$:

$(102)^5 = 10,000,000,000 + 1,000,000,000 + 40,000,000 + 800,000 + 8000 + 32$

$(102)^5 = 11,000,000,000 + 40,000,000 + 800,000 + 8000 + 32$

$(102)^5 = 11,040,000,000 + 800,000 + 8000 + 32$

$(102)^5 = 11,040,800,000 + 8000 + 32$

$(102)^5 = 11,040,808,000 + 32$

$(102)^5 = 11,040,808,032$

The value of $(102)^5$ evaluated using the binomial theorem is $11,040,808,032$.

Solution:

We need to evaluate $(101)^4$ using the binomial theorem.

We can write 101 as $100 + 1$. So, $(101)^4 = (100 + 1)^4$.

This is in the form of $(a+b)^n$, where $a=100$, $b=1$, and $n=4$.

We use the Binomial Theorem, which states that for a positive integer $n$,

$(a+b)^n = \sum\limits_{r=0}^{n} \binom{n}{r} a^{n-r} b^{r}$

For $n=4$, the expansion is:

$(100+1)^4 = \binom{4}{0}(100)^4(1)^0 + \binom{4}{1}(100)^3(1)^1 + \binom{4}{2}(100)^2(1)^2 + \binom{4}{3}(100)^1(1)^3 + \binom{4}{4}(100)^0(1)^4$

Now, we calculate each term using the binomial coefficients: $\binom{4}{0}=1, \binom{4}{1}=4, \binom{4}{2}=6, \binom{4}{3}=4, \binom{4}{4}=1$.

Term for $r=0$: $\binom{4}{0}(100)^4(1)^0 = 1 \cdot (100,000,000) \cdot 1 = 100,000,000$

Term for $r=1$: $\binom{4}{1}(100)^3(1)^1 = 4 \cdot (1,000,000) \cdot 1 = 4,000,000$

Term for $r=2$: $\binom{4}{2}(100)^2(1)^2 = 6 \cdot (10,000) \cdot 1 = 60,000$

Term for $r=3$: $\binom{4}{3}(100)^1(1)^3 = 4 \cdot (100) \cdot 1 = 400$

Term for $r=4$: $\binom{4}{4}(100)^0(1)^4 = 1 \cdot 1 \cdot 1 = 1$

Summing these terms, we get the value of $(101)^4$:

$(101)^4 = 100,000,000 + 4,000,000 + 60,000 + 400 + 1$

$(101)^4 = 104,000,000 + 60,000 + 400 + 1$

$(101)^4 = 104,060,000 + 400 + 1$

$(101)^4 = 104,060,400 + 1$

$(101)^4 = 104,060,401$

The value of $(101)^4$ evaluated using the binomial theorem is $104,060,401$.

Solution:

We need to evaluate $(99)^5$ using the binomial theorem.

We can write 99 as $100 - 1$. So, $(99)^5 = (100 - 1)^5$.

This is in the form of $(a-b)^n$, where $a=100$, $b=1$, and $n=5$.

We use the Binomial Theorem for $(a-b)^n$, which states:

$(a-b)^n = \sum\limits_{r=0}^{n} \binom{n}{r} a^{n-r} (-b)^{r}$

For $n=5$, the expansion is:

$(100-1)^5 = \binom{5}{0}(100)^5(-1)^0 + \binom{5}{1}(100)^4(-1)^1 + \binom{5}{2}(100)^3(-1)^2 + \binom{5}{3}(100)^2(-1)^3 + \binom{5}{4}(100)^1(-1)^4 + \binom{5}{5}(100)^0(-1)^5$

Now, we calculate each term using the binomial coefficients: $\binom{5}{0}=1, \binom{5}{1}=5, \binom{5}{2}=10, \binom{5}{3}=10, \binom{5}{4}=5, \binom{5}{5}=1$.

Term for $r=0$: $\binom{5}{0}(100)^5(-1)^0 = 1 \cdot (10000000000) \cdot 1 = 10,000,000,000$

Term for $r=1$: $\binom{5}{1}(100)^4(-1)^1 = 5 \cdot (100000000) \cdot (-1) = -500,000,000$

Term for $r=2$: $\binom{5}{2}(100)^3(-1)^2 = 10 \cdot (1000000) \cdot 1 = 10,000,000$

Term for $r=3$: $\binom{5}{3}(100)^2(-1)^3 = 10 \cdot (10000) \cdot (-1) = -100,000$

Term for $r=4$: $\binom{5}{4}(100)^1(-1)^4 = 5 \cdot (100) \cdot 1 = 500$

Term for $r=5$: $\binom{5}{5}(100)^0(-1)^5 = 1 \cdot 1 \cdot (-1) = -1$

Summing these terms, we get the value of $(99)^5$:

$(99)^5 = 10,000,000,000 + (-500,000,000) + 10,000,000 + (-100,000) + 500 + (-1)$

$(99)^5 = 10,000,000,000 - 500,000,000 + 10,000,000 - 100,000 + 500 - 1$

$(99)^5 = 9,500,000,000 + 10,000,000 - 100,000 + 500 - 1$

$(99)^5 = 9,510,000,000 - 100,000 + 500 - 1$

$(99)^5 = 9,509,900,000 + 500 - 1$

$(99)^5 = 9,509,900,500 - 1$

$(99)^5 = 9,509,900,499$

The value of $(99)^5$ evaluated using the binomial theorem is $9,509,900,499$.

Solution:

We need to compare $(1.1)^{10000}$ and $1000$.

Consider the expression $(1.1)^{10000}$. We can write $1.1$ as $1 + 0.1$. So, we need to evaluate $(1 + 0.1)^{10000}$.

This is in the form $(1+x)^n$, where $x = 0.1$ and $n = 10000$. We can use the Binomial Theorem to expand this expression.

The Binomial Theorem for $(1+x)^n$, where $n$ is a positive integer, is given by:

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

Substituting $n = 10000$ and $x = 0.1$, we get:

$(1 + 0.1)^{10000} = \binom{10000}{0} + \binom{10000}{1}(0.1) + \binom{10000}{2}(0.1)^2 + \binom{10000}{3}(0.1)^3 + \dots + \binom{10000}{10000}(0.1)^{10000}$

Let's calculate the first two terms of the expansion:

The first term is $\binom{10000}{0} = 1$.

The second term is $\binom{10000}{1}(0.1) = 10000 \cdot (0.1) = 10000 \cdot \frac{1}{10} = 1000$.

So, the expansion begins as:

$(1.1)^{10000} = 1 + 1000 + \binom{10000}{2}(0.1)^2 + \binom{10000}{3}(0.1)^3 + \dots + \binom{10000}{10000}(0.1)^{10000}$

$(1.1)^{10000} = 1001 + \binom{10000}{2}(0.1)^2 + \binom{10000}{3}(0.1)^3 + \dots + \binom{10000}{10000}(0.1)^{10000}$

Since $n=10000$ is a positive integer and $x=0.1$ is positive, all the terms $\binom{n}{r}x^r$ for $r \ge 2$ in the expansion are positive.

Therefore, the sum of the remaining terms, $\binom{10000}{2}(0.1)^2 + \binom{10000}{3}(0.1)^3 + \dots$, is a positive value.

This means $(1.1)^{10000} = 1001 + (\text{a positive value})$.

Hence, $(1.1)^{10000} > 1001$.

We are comparing $(1.1)^{10000}$ with $1000$.

We have shown that $(1.1)^{10000} > 1001$.

Since $1001 > 1000$, by the transitive property of inequality, we can conclude that $(1.1)^{10000}$ is greater than $1000$.

Thus, $(1.1)^{10000} > 1000$.

Conclusion: $(1.1)^{10000}$ is larger than $1000$.

Solution:

First, we need to find the expansion of $(a + b)^4 – (a – b)^4$.

We will use the Binomial Theorem to expand $(a+b)^4$ and $(a-b)^4$ separately.

The Binomial Theorem states that $(x+y)^n = \sum\limits_{r=0}^{n} \binom{n}{r} x^{n-r} y^{r}$.

For $(a+b)^4$, we have $n=4$, $x=a$, and $y=b$:

$(a+b)^4 = \binom{4}{0}a^4b^0 + \binom{4}{1}a^3b^1 + \binom{4}{2}a^2b^2 + \binom{4}{3}a^1b^3 + \binom{4}{4}a^0b^4$

Calculating the binomial coefficients: $\binom{4}{0}=1, \binom{4}{1}=4, \binom{4}{2}=6, \binom{4}{3}=4, \binom{4}{4}=1$.

So, $(a+b)^4 = 1 \cdot a^4 \cdot 1 + 4 \cdot a^3 \cdot b + 6 \cdot a^2 \cdot b^2 + 4 \cdot a \cdot b^3 + 1 \cdot 1 \cdot b^4$

$(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$

For $(a-b)^4$, we have $n=4$, $x=a$, and $y=-b$:

$(a-b)^4 = \binom{4}{0}a^4(-b)^0 + \binom{4}{1}a^3(-b)^1 + \binom{4}{2}a^2(-b)^2 + \binom{4}{3}a^1(-b)^3 + \binom{4}{4}a^0(-b)^4$

$(a-b)^4 = 1 \cdot a^4 \cdot 1 + 4 \cdot a^3 \cdot (-b) + 6 \cdot a^2 \cdot b^2 + 4 \cdot a \cdot (-b^3) + 1 \cdot 1 \cdot b^4$

$(a-b)^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4$

Now, we subtract the expansion of $(a-b)^4$ from the expansion of $(a+b)^4$:

$(a+b)^4 - (a-b)^4 = (a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4) - (a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4)$

$(a+b)^4 - (a-b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 - a^4 + 4a^3b - 6a^2b^2 + 4ab^3 - b^4$

Combine like terms:

$= (a^4 - a^4) + (4a^3b + 4a^3b) + (6a^2b^2 - 6a^2b^2) + (4ab^3 + 4ab^3) + (b^4 - b^4)$

$= 0 + 8a^3b + 0 + 8ab^3 + 0$

$(a+b)^4 - (a-b)^4 = 8a^3b + 8ab^3$

We can factor out $8ab$ from the result:

$(a+b)^4 - (a-b)^4 = 8ab(a^2 + b^2)$

Next, we need to evaluate $(\sqrt{3} + \sqrt{2})^4 - (\sqrt{3} - \sqrt{2})^4$ using the result we just found.

Comparing this expression with $(a+b)^4 - (a-b)^4$, we can identify $a = \sqrt{3}$ and $b = \sqrt{2}$.

Substitute these values into the simplified expression $8ab(a^2 + b^2)$.

First, calculate $a^2$ and $b^2$:

$a^2 = (\sqrt{3})^2 = 3$

$b^2 = (\sqrt{2})^2 = 2$

Now substitute $a=\sqrt{3}$, $b=\sqrt{2}$, $a^2=3$, and $b^2=2$ into the expression $8ab(a^2 + b^2)$:

$(\sqrt{3} + \sqrt{2})^4 - (\sqrt{3} - \sqrt{2})^4 = 8(\sqrt{3})(\sqrt{2})((\sqrt{3})^2 + (\sqrt{2})^2)$

$= 8\sqrt{3 \cdot 2}(3 + 2)$

$= 8\sqrt{6}(5)$

$= 40\sqrt{6}$

Thus, the value of $(\sqrt{3} + \sqrt{2})^4 - (\sqrt{3} - \sqrt{2})^4$ is $40\sqrt{6}$.

Solution:

Part 1: Find $(x + 1)^6 + (x – 1)^6$

We use the Binomial Theorem to expand $(x+1)^6$ and $(x-1)^6$. The Binomial Theorem for $(a+b)^n$ is given by $\sum\limits_{r=0}^{n} \binom{n}{r} a^{n-r} b^{r}$.

Expansion of $(x+1)^6$:

Here $a=x, b=1, n=6$. The terms are:

$\binom{6}{0}x^6(1)^0 = 1 \cdot x^6 \cdot 1 = x^6$

$\binom{6}{1}x^5(1)^1 = 6 \cdot x^5 \cdot 1 = 6x^5$

$\binom{6}{2}x^4(1)^2 = 15 \cdot x^4 \cdot 1 = 15x^4$

$\binom{6}{3}x^3(1)^3 = 20 \cdot x^3 \cdot 1 = 20x^3$

$\binom{6}{4}x^2(1)^4 = 15 \cdot x^2 \cdot 1 = 15x^2$

$\binom{6}{5}x^1(1)^5 = 6 \cdot x^1 \cdot 1 = 6x$

$\binom{6}{6}x^0(1)^6 = 1 \cdot 1 \cdot 1 = 1$

So, $(x+1)^6 = x^6 + 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1$

Expansion of $(x-1)^6$:

Here $a=x, b=-1, n=6$. The terms will have alternating signs:

$\binom{6}{0}x^6(-1)^0 = 1 \cdot x^6 \cdot 1 = x^6$

$\binom{6}{1}x^5(-1)^1 = 6 \cdot x^5 \cdot (-1) = -6x^5$

$\binom{6}{2}x^4(-1)^2 = 15 \cdot x^4 \cdot 1 = 15x^4$

$\binom{6}{3}x^3(-1)^3 = 20 \cdot x^3 \cdot (-1) = -20x^3$

$\binom{6}{4}x^2(-1)^4 = 15 \cdot x^2 \cdot 1 = 15x^2$

$\binom{6}{5}x^1(-1)^5 = 6 \cdot x^1 \cdot (-1) = -6x$

$\binom{6}{6}x^0(-1)^6 = 1 \cdot 1 \cdot 1 = 1$

So, $(x-1)^6 = x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1$

Now, we add the two expansions:

$(x + 1)^6 + (x – 1)^6 = (x^6 + 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1) + (x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1)$

Combine the like terms:

$(x + 1)^6 + (x – 1)^6 = (x^6 + x^6) + (6x^5 - 6x^5) + (15x^4 + 15x^4) + (20x^3 - 20x^3) + (15x^2 + 15x^2) + (6x - 6x) + (1 + 1)$

$(x + 1)^6 + (x – 1)^6 = 2x^6 + 0 + 30x^4 + 0 + 30x^2 + 0 + 2$

$(x + 1)^6 + (x – 1)^6 = 2x^6 + 30x^4 + 30x^2 + 2$

Part 2: Evaluate $(\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6$

We can use the result from Part 1 by substituting $x = \sqrt{2}$ into the expression $2x^6 + 30x^4 + 30x^2 + 2$.

If $x = \sqrt{2}$, then:

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

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

$x^6 = (x^2)^3 = (2)^3 = 8$

Substitute these values into $2x^6 + 30x^4 + 30x^2 + 2$:

$2(8) + 30(4) + 30(2) + 2$

$= 16 + 120 + 60 + 2$

$= 136 + 60 + 2$

$= 196 + 2$

$= 198$

The value of $(x + 1)^6 + (x – 1)^6$ is $2x^6 + 30x^4 + 30x^2 + 2$.

The value of $(\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6$ is $198$.

Given:

The expression $9^{n+1} – 8n – 9$, where $n$ is a positive integer.

To Prove:

That $9^{n+1} – 8n – 9$ is divisible by 64 whenever $n$ is a positive integer.

Proof:

Consider the term $9^{n+1}$. We can write $9$ as $1+8$. Thus, $9^{n+1} = (1+8)^{n+1}$.

Using the Binomial Theorem, the expansion of $(a+b)^m$ for a positive integer $m$ is given by:

$(a+b)^m = \sum\limits_{r=0}^{m} \binom{m}{r} a^{m-r} b^r$

Let $a=1$, $b=8$, and $m=n+1$. Expanding $(1+8)^{n+1}$ using the Binomial Theorem:

$(1+8)^{n+1} = \sum\limits_{r=0}^{n+1} \binom{n+1}{r} 1^{(n+1)-r} 8^r$

$(1+8)^{n+1} = \binom{n+1}{0}1^{n+1}8^0 + \binom{n+1}{1}1^{n}8^1 + \binom{n+1}{2}1^{n-1}8^2 + \binom{n+1}{3}1^{n-2}8^3 + \dots + \binom{n+1}{n+1}1^0 8^{n+1}$

Simplifying the terms, knowing that $\binom{n+1}{0} = 1$ and $\binom{n+1}{1} = n+1$:

$9^{n+1} = 1 + 8(n+1) + \binom{n+1}{2}8^2 + \binom{n+1}{3}8^3 + \dots + 8^{n+1}$

$9^{n+1} = 1 + 8n + 8 + \binom{n+1}{2}8^2 + \binom{n+1}{3}8^3 + \dots + 8^{n+1}$

$9^{n+1} = 9 + 8n + \binom{n+1}{2}8^2 + \binom{n+1}{3}8^3 + \dots + 8^{n+1}$

Now, consider the expression $9^{n+1} - 8n - 9$. Substitute the expansion of $9^{n+1}$ into this expression:

$9^{n+1} - 8n - 9 = \left(9 + 8n + \binom{n+1}{2}8^2 + \binom{n+1}{3}8^3 + \dots + 8^{n+1}\right) - 8n - 9$

$9^{n+1} - 8n - 9 = 9 + 8n + \binom{n+1}{2}8^2 + \binom{n+1}{3}8^3 + \dots + 8^{n+1} - 8n - 9$

Combine the constant terms and the terms involving $n$:

$9^{n+1} - 8n - 9 = (9 - 9) + (8n - 8n) + \binom{n+1}{2}8^2 + \binom{n+1}{3}8^3 + \dots + 8^{n+1}$

$9^{n+1} - 8n - 9 = 0 + 0 + \binom{n+1}{2}8^2 + \binom{n+1}{3}8^3 + \dots + 8^{n+1}$

$9^{n+1} - 8n - 9 = \binom{n+1}{2}8^2 + \binom{n+1}{3}8^3 + \dots + \binom{n+1}{n+1}8^{n+1}$

We can express this sum using summation notation:

$9^{n+1} - 8n - 9 = \sum\limits_{r=2}^{n+1} \binom{n+1}{r} 8^r$

For any integer $r \ge 2$, we can write $8^r$ as $8^2 \cdot 8^{r-2}$. Since $8^2 = 64$, we have $8^r = 64 \cdot 8^{r-2}$.

Substitute this into the sum:

$9^{n+1} - 8n - 9 = \sum\limits_{r=2}^{n+1} \binom{n+1}{r} (64 \cdot 8^{r-2})$

Factor out the constant 64 from the sum:

$9^{n+1} - 8n - 9 = 64 \left(\sum\limits_{r=2}^{n+1} \binom{n+1}{r} 8^{r-2}\right)$

Let $K = \sum\limits_{r=2}^{n+1} \binom{n+1}{r} 8^{r-2}$.

Since $n$ is a positive integer, $n \in \{1, 2, 3, \dots\}$. This implies $n+1 \ge 2$. The summation runs from $r=2$ up to $n+1$. For each integer $r$ in this range, the binomial coefficient $\binom{n+1}{r}$ is an integer. The term $8^{r-2}$ is also an integer since $r \ge 2$ means $r-2 \ge 0$. The product of integers is an integer, and the sum of integers is an integer. Therefore, $K$ is an integer for all positive integers $n$.

This equation shows that $9^{n+1} - 8n - 9$ can be expressed as 64 multiplied by an integer $K$. This means $9^{n+1} - 8n - 9$ is a multiple of 64.

Conclusion:

Using the Binomial Theorem, we have proven that $9^{n+1} – 8n – 9$ is always a multiple of 64 for any positive integer $n$. Therefore, $9^{n+1} – 8n – 9$ is divisible by 64.

To Prove:

$\sum\limits_{r=0}^{n} 3^r\; ^nC_r = 4^n$

Proof:

We will use the Binomial Theorem, which states that for any real numbers $x$ and $y$, and any non-negative integer $n$:

$(x+y)^n = \sum\limits_{r=0}^{n} \binom{n}{r} x^{n-r} y^r$

or equivalently,

$(x+y)^n = \sum\limits_{r=0}^{n} \binom{n}{r} x^{r} y^{n-r}$

The given summation is $\sum\limits_{r=0}^{n} 3^r\; ^nC_r$. Recall that $^nC_r = \binom{n}{r}$. So the sum is $\sum\limits_{r=0}^{n} \binom{n}{r} 3^r$.

Let's compare this sum with the binomial expansion formula $(x+y)^n = \sum\limits_{r=0}^{n} \binom{n}{r} x^{n-r} y^r$.

We need to match the term $\binom{n}{r} 3^r$ with $\binom{n}{r} x^{n-r} y^r$.

If we choose $y = 3$, the term becomes $\binom{n}{r} x^{n-r} 3^r$.

For this to match $\binom{n}{r} 3^r$, we must have $x^{n-r} = 1$ for all values of $r$ from $0$ to $n$. This is true if we choose $x = 1$.

Let $x = 1$ and $y = 3$ in the binomial expansion of $(x+y)^n$:

$(1+3)^n = \sum\limits_{r=0}^{n} \binom{n}{r} 1^{n-r} 3^r$

Since $1^{n-r} = 1$ for any integer value of $n-r$, the right side simplifies to:

$\sum\limits_{r=0}^{n} \binom{n}{r} 1 \cdot 3^r = \sum\limits_{r=0}^{n} \binom{n}{r} 3^r = \sum\limits_{r=0}^{n} 3^r\; ^nC_r$

The left side of the equation is $(1+3)^n = 4^n$.

Therefore, we have:

$4^n = \sum\limits_{r=0}^{n} 3^r\; ^nC_r$

This proves the required identity.

Hence, $\sum\limits_{r=0}^{n} 3^r\; ^nC_r = 4^n$.

Given:

a and b are distinct integers, and n is a positive integer.

To Prove:

$(a - b)$ is a factor of $a^n - b^n$.

Proof:

We want to show that $a^n - b^n$ can be expressed as the product of $(a-b)$ and some integer $K$.

Following the hint, we can rewrite $a^n$ by adding and subtracting b within the base:

$a^n = ((a - b) + b)^n$

Now, we apply the binomial theorem to expand $((a - b) + b)^n$. Let $x = (a-b)$ and $y = b$. Then $(x+y)^n = \sum\limits_{k=0}^n \binom{n}{k}x^{n-k}y^k$. Or in the form used in the steps below, $(x+y)^n = \sum\limits_{k=0}^n \binom{n}{k}x^k y^{n-k}$.

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

We know that $\binom{n}{0} = 1$ and $(a-b)^0 = 1$ (since a and b are distinct, $a-b \neq 0$). Also, $\binom{n}{n} = 1$ and $b^0 = 1$. So, the expansion becomes:

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

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

Now, subtract $b^n$ from both sides of the equation:

$a^n - b^n = \left(b^n + \binom{n}{1}(a-b)b^{n-1} + \binom{n}{2}(a-b)^2 b^{n-2} + ... + (a-b)^n\right) - b^n$

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

Observe that every term on the right-hand side of the equation contains $(a-b)$ as a factor (since n is a positive integer, $k$ ranges from 1 to n, so $(a-b)^k$ for $k \ge 1$ has a factor of $(a-b)$).

We can factor out $(a-b)$ from the entire expression:

$a^n - b^n = (a-b) \left[ \binom{n}{1}b^{n-1} + \binom{n}{2}(a-b)b^{n-2} + ... + \binom{n}{n-1}(a-b)^{n-2}b + (a-b)^{n-1} \right]$

Let $K = \binom{n}{1}b^{n-1} + \binom{n}{2}(a-b)b^{n-2} + ... + \binom{n}{n-1}(a-b)^{n-2}b + (a-b)^{n-1}$.

Since a and b are integers, $(a-b)$ is an integer. For any positive integer n and $1 \le k \le n$, the binomial coefficients $\binom{n}{k}$ are integers. The terms $(a-b)^{k-1}$ and $b^{n-k}$ (for $k \ge 1$) are also integers because products and powers of integers are integers.

Therefore, the expression inside the square brackets, $K$, which is a sum of products of integers, is also an integer.

We have shown that $a^n - b^n = (a-b) \cdot K$, where $K$ is an integer.

This implies that $a^n - b^n$ is a multiple of $(a-b)$.

Hence, $(a-b)$ is a factor of $a^n - b^n$ for any positive integer $n$.

This completes the proof.

Given:

The expression to evaluate is $(\sqrt{3} + \sqrt{2})^6 - (\sqrt{3} - \sqrt{2})^6$.

To Evaluate:

We need to find the numerical value of the given expression.

Solution:

We can use the binomial theorem to expand the terms $(\sqrt{3} + \sqrt{2})^6$ and $(\sqrt{3} - \sqrt{2})^6$.

Recall the binomial expansion formula: $(a+b)^n = \sum\limits_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$ and $(a-b)^n = \sum\limits_{k=0}^{n} \binom{n}{k} a^{n-k} (-b)^k$.

Let $a = \sqrt{3}$ and $b = \sqrt{2}$. The expression is of the form $(a+b)^6 - (a-b)^6$.

Expanding $(a+b)^6$:

$(a+b)^6 = \binom{6}{0}a^6 + \binom{6}{1}a^5b + \binom{6}{2}a^4b^2 + \binom{6}{3}a^3b^3 + \binom{6}{4}a^2b^4 + \binom{6}{5}ab^5 + \binom{6}{6}b^6$

Expanding $(a-b)^6$:

$(a-b)^6 = \binom{6}{0}a^6 - \binom{6}{1}a^5b + \binom{6}{2}a^4b^2 - \binom{6}{3}a^3b^3 + \binom{6}{4}a^2b^4 - \binom{6}{5}ab^5 + \binom{6}{6}b^6$

Now, let's find the difference:

$(a+b)^6 - (a-b)^6 = \left(\binom{6}{0}a^6 + \binom{6}{1}a^5b + ... + \binom{6}{6}b^6\right) - \left(\binom{6}{0}a^6 - \binom{6}{1}a^5b - ... + \binom{6}{6}b^6\right)$

When subtracting, the terms with even powers of $b$ cancel out, and the terms with odd powers of $b$ are doubled:

For $n=6$, $a=\sqrt{3}$, and $b=\sqrt{2}$, the expression becomes:

$(\sqrt{3} + \sqrt{2})^6 - (\sqrt{3} - \sqrt{2})^6 = 2\left[\binom{6}{1}(\sqrt{3})^{6-1}(\sqrt{2})^1 + \binom{6}{3}(\sqrt{3})^{6-3}(\sqrt{2})^3 + \binom{6}{5}(\sqrt{3})^{6-5}(\sqrt{2})^5 \right]$

$= 2\left[\binom{6}{1}(\sqrt{3})^5(\sqrt{2})^1 + \binom{6}{3}(\sqrt{3})^3(\sqrt{2})^3 + \binom{6}{5}(\sqrt{3})^1(\sqrt{2})^5 \right]$

Let's calculate the required binomial coefficients:

$\binom{6}{1} = 6$

$\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$

$\binom{6}{5} = \binom{6}{6-5} = \binom{6}{1} = 6$

Now, let's calculate the powers of $\sqrt{3}$ and $\sqrt{2}$:

$(\sqrt{3})^5 = (\sqrt{3})^4 \cdot \sqrt{3} = (3^2) \cdot \sqrt{3} = 9\sqrt{3}$

$(\sqrt{3})^3 = (\sqrt{3})^2 \cdot \sqrt{3} = 3\sqrt{3}$

$(\sqrt{3})^1 = \sqrt{3}$

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

$(\sqrt{2})^3 = (\sqrt{2})^2 \cdot \sqrt{2} = 2\sqrt{2}$

$(\sqrt{2})^5 = (\sqrt{2})^4 \cdot \sqrt{2} = (2^2) \cdot \sqrt{2} = 4\sqrt{2}$

Substitute these values back into the expression:

$(\sqrt{3} + \sqrt{2})^6 - (\sqrt{3} - \sqrt{2})^6 = 2\left[6(9\sqrt{3})(\sqrt{2}) + 20(3\sqrt{3})(2\sqrt{2}) + 6(\sqrt{3})(4\sqrt{2})\right]$

Simplify each term inside the bracket:

$6(9\sqrt{3})(\sqrt{2}) = 54\sqrt{3 \times 2} = 54\sqrt{6}$

$20(3\sqrt{3})(2\sqrt{2}) = 20(3 \times 2)(\sqrt{3}\sqrt{2}) = 20(6)\sqrt{6} = 120\sqrt{6}$

$6(\sqrt{3})(4\sqrt{2}) = 6(4)(\sqrt{3}\sqrt{2}) = 24\sqrt{6}$

Sum the terms inside the bracket:

$54\sqrt{6} + 120\sqrt{6} + 24\sqrt{6} = (54 + 120 + 24)\sqrt{6} = 198\sqrt{6}$

Now, multiply the sum by 2:

$2 \times 198\sqrt{6} = 396\sqrt{6}$

The value of the expression $(\sqrt{3} + \sqrt{2})^6 - (\sqrt{3} - \sqrt{2})^6$ is $396\sqrt{6}$.

Let the given expression be $E$.

$E = (a^2 + \sqrt{a^2 - 1})^4 + (a^2 - \sqrt{a^2 - 1})^4$

To simplify the expression, let's use substitution.

Let $x = a^2$ and $y = \sqrt{a^2 - 1}$.

Then the expression can be written as:

$E = (x + y)^4 + (x - y)^4$

We use the binomial expansion formula $(m+n)^4 = m^4 + 4m^3n + 6m^2n^2 + 4mn^3 + n^4$.

Expanding $(x+y)^4$:

$(x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$

Expanding $(x-y)^4$:

$(x-y)^4 = x^4 - 4x^3y + 6x^2y^2 - 4xy^3 + y^4$

Now, add the two expansions:

$(x+y)^4 + (x-y)^4 = (x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4) + (x^4 - 4x^3y + 6x^2y^2 - 4xy^3 + y^4)$

Combine the like terms. The terms with odd powers of $y$ ($4x^3y$ and $4xy^3$) cancel out:

$(x+y)^4 + (x-y)^4 = x^4 + x^4 + 6x^2y^2 + 6x^2y^2 + y^4 + y^4$

$(x+y)^4 + (x-y)^4 = 2x^4 + 12x^2y^2 + 2y^4$

Factor out $2$ from the expression:

$(x+y)^4 + (x-y)^4 = 2(x^4 + 6x^2y^2 + y^4)$

Now, substitute back $x = a^2$ and $y = \sqrt{a^2 - 1}$ into the simplified expression $2(x^4 + 6x^2y^2 + y^4)$.

First, let's find the required powers of $x$ and $y$ in terms of $a$:

$x = a^2 \implies x^2 = (a^2)^2 = a^4 \implies x^4 = (a^2)^4 = a^8$

$y = \sqrt{a^2 - 1} \implies y^2 = (\sqrt{a^2 - 1})^2 = a^2 - 1$

$y^4 = (y^2)^2 = (a^2 - 1)^2$

Expand $(a^2 - 1)^2$ using the formula $(m-n)^2 = m^2 - 2mn + n^2$:

$(a^2 - 1)^2 = (a^2)^2 - 2(a^2)(1) + 1^2 = a^4 - 2a^2 + 1$

So, $y^4 = a^4 - 2a^2 + 1$.

Substitute these values back into $2(x^4 + 6x^2y^2 + y^4)$:

$E = 2(a^8 + 6(a^4)(a^2 - 1) + (a^4 - 2a^2 + 1))$

Now, simplify the expression inside the parenthesis:

$E = 2(a^8 + 6a^4 \times a^2 - 6a^4 \times 1 + a^4 - 2a^2 + 1)$

$E = 2(a^8 + 6a^6 - 6a^4 + a^4 - 2a^2 + 1)$

Combine the terms with $a^4$:

$E = 2(a^8 + 6a^6 + (-6 + 1)a^4 - 2a^2 + 1)$

$E = 2(a^8 + 6a^6 - 5a^4 - 2a^2 + 1)$

Finally, distribute the $2$ to each term inside the parenthesis:

$E = 2 \times a^8 + 2 \times 6a^6 - 2 \times 5a^4 - 2 \times 2a^2 + 2 \times 1$

$E = 2a^8 + 12a^6 - 10a^4 - 4a^2 + 2$

The value of the expression is $2a^8 + 12a^6 - 10a^4 - 4a^2 + 2$.

Given:

The expression is $(0.99)^5$.

We need to find an approximation using the first three terms of its expansion.

To Find:

An approximation of $(0.99)^5$ using the first three terms of its binomial expansion.

Solution:

We can write $0.99$ as $1 - 0.01$. So, the expression becomes $(1 - 0.01)^5$.

This is in the form of $(1+x)^n$, where $x = -0.01$ and $n = 5$.

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

$(1+x)^n = ^nC_0 + ^nC_1 x + ^nC_2 x^2 + ^nC_3 x^3 + ... + ^nC_n x^n$

We need to use the first three terms of the expansion, which are $^nC_0$, $^nC_1 x$, and $^nC_2 x^2$.

For $(1 - 0.01)^5$, we have $n=5$ and $x=-0.01$.

The first term is $^5C_0$:

$^5C_0 = 1$

The second term is $^5C_1 x$:

$^5C_1 = 5$

Second term = $^5C_1 \times (-0.01) = 5 \times (-0.01) = -0.05$

The third term is $^5C_2 x^2$:

$^5C_2 = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10$

Third term = $^5C_2 \times (-0.01)^2 = 10 \times (0.0001) = 0.0010$

To find the approximation, we sum the first three terms:

Approximation $\approx$ First Term + Second Term + Third Term

Approximation $\approx 1 + (-0.05) + 0.0010$

Approximation $\approx 1 - 0.05 + 0.0010$

Approximation $\approx 0.95 + 0.0010$

Approximation $\approx 0.9510$

The approximation of $(0.99)^5$ using the first three terms of its expansion is $0.9510$.

Given:

The expression to expand is $\left(1+ \frac{x}{2} -\frac{2}{x}\right)^{4}$, where $x \neq 0$.

To Expand:

Expand $\left(1+ \frac{x}{2} -\frac{2}{x}\right)^{4}$ using the Binomial Theorem.

Solution:

We can group the terms inside the parenthesis to form a binomial. Let $y = \frac{x}{2} - \frac{2}{x}$. The expression becomes $(1+y)^4$.

Using the Binomial Theorem for $(1+y)^4$, we have:

$(1+y)^4 = ^4C_0 (1)^4 y^0 + ^4C_1 (1)^3 y^1 + ^4C_2 (1)^2 y^2 + ^4C_3 (1)^1 y^3 + ^4C_4 (1)^0 y^4$

$(1+y)^4 = 1 \cdot 1 \cdot y^0 + 4 \cdot 1 \cdot y^1 + 6 \cdot 1 \cdot y^2 + 4 \cdot 1 \cdot y^3 + 1 \cdot 1 \cdot y^4$

$(1+y)^4 = 1 + 4y + 6y^2 + 4y^3 + y^4$

Now, we substitute $y = \frac{x}{2} - \frac{2}{x}$ back into this expansion and calculate the powers of $y$:

$y = \frac{x}{2} - \frac{2}{x}$

$y^2 = \left(\frac{x}{2} - \frac{2}{x}\right)^2 = \left(\frac{x}{2}\right)^2 - 2\left(\frac{x}{2}\right)\left(\frac{2}{x}\right) + \left(\frac{2}{x}\right)^2 = \frac{x^2}{4} - 2 + \frac{4}{x^2}$

$y^3 = \left(\frac{x}{2} - \frac{2}{x}\right)^3 = \left(\frac{x}{2}\right)^3 - 3\left(\frac{x}{2}\right)^2\left(\frac{2}{x}\right) + 3\left(\frac{x}{2}\right)\left(\frac{2}{x}\right)^2 - \left(\frac{2}{x}\right)^3$

$y^3 = \frac{x^3}{8} - 3\left(\frac{x^2}{4}\right)\left(\frac{2}{x}\right) + 3\left(\frac{x}{2}\right)\left(\frac{4}{x^2}\right) - \frac{8}{x^3}$

$y^3 = \frac{x^3}{8} - \frac{6x^2}{4x} + \frac{12x}{2x^2} - \frac{8}{x^3} = \frac{x^3}{8} - \frac{3x}{2} + \frac{6}{x} - \frac{8}{x^3}$

$y^4 = (y^2)^2 = \left(\frac{x^2}{4} - 2 + \frac{4}{x^2}\right)^2$

$y^4 = \left(\frac{x^2}{4}\right)^2 + (-2)^2 + \left(\frac{4}{x^2}\right)^2 + 2\left(\frac{x^2}{4}\right)(-2) + 2\left(\frac{x^2}{4}\right)\left(\frac{4}{x^2}\right) + 2(-2)\left(\frac{4}{x^2}\right)$

$y^4 = \frac{x^4}{16} + 4 + \frac{16}{x^4} - x^2 + 2 - \frac{16}{x^2} = \frac{x^4}{16} - x^2 - \frac{16}{x^2} + \frac{16}{x^4} + 6$

Now, substitute these expressions for $y, y^2, y^3, y^4$ into the expansion $1 + 4y + 6y^2 + 4y^3 + y^4$:

$1 + 4\left(\frac{x}{2} - \frac{2}{x}\right) + 6\left(\frac{x^2}{4} - 2 + \frac{4}{x^2}\right) + 4\left(\frac{x^3}{8} - \frac{3x}{2} + \frac{6}{x} - \frac{8}{x^3}\right) + \left(\frac{x^4}{16} - x^2 - \frac{16}{x^2} + \frac{16}{x^4} + 6\right)$

Expand each term:

$1 + \left(2x - \frac{8}{x}\right) + \left(\frac{3x^2}{2} - 12 + \frac{24}{x^2}\right) + \left(\frac{x^3}{2} - 6x + \frac{24}{x} - \frac{32}{x^3}\right) + \left(\frac{x^4}{16} - x^2 - \frac{16}{x^2} + \frac{16}{x^4} + 6\right)$

Group and combine like terms (terms with the same power of $x$):

Constant terms: $1 - 12 + 6 = -5$

Terms with $x$: $2x - 6x = -4x$

Terms with $\frac{1}{x}$: $-\frac{8}{x} + \frac{24}{x} = \frac{16}{x}$

Terms with $x^2$: $\frac{3x^2}{2} - x^2 = \frac{3x^2}{2} - \frac{2x^2}{2} = \frac{x^2}{2}$

Terms with $\frac{1}{x^2}$: $\frac{24}{x^2} - \frac{16}{x^2} = \frac{8}{x^2}$

Terms with $x^3$: $\frac{x^3}{2}$

Terms with $\frac{1}{x^3}$: $-\frac{32}{x^3}$

Terms with $x^4$: $\frac{x^4}{16}$

Terms with $\frac{1}{x^4}$: $\frac{16}{x^4}$

Combine all the combined terms:

$\frac{x^4}{16} + \frac{x^3}{2} + \frac{x^2}{2} - 4x - 5 + \frac{16}{x} + \frac{8}{x^2} - \frac{32}{x^3} + \frac{16}{x^4}$

The expansion of $\left(1+ \frac{x}{2} -\frac{2}{x}\right)^{4}$ is $\frac{x^4}{16} + \frac{x^3}{2} + \frac{x^2}{2} - 4x - 5 + \frac{16}{x} + \frac{8}{x^2} - \frac{32}{x^3} + \frac{16}{x^4}$.

Given:

The expression is $(3x^2 – 2ax + 3a^2)^3$.

To Expand:

Expand $(3x^2 – 2ax + 3a^2)^3$ using the Binomial Theorem.

Solution:

To use the Binomial Theorem, we group the terms of the trinomial to form a binomial. Let's group the first and third terms together:

$(3x^2 – 2ax + 3a^2)^3 = ((3x^2 + 3a^2) - 2ax)^3$

Let $A = 3x^2 + 3a^2$ and $B = -2ax$. The expression becomes $(A+B)^3$.

Using the Binomial Theorem for $(A+B)^3$:

$(A+B)^3 = ^3C_0 A^3 B^0 + ^3C_1 A^2 B^1 + ^3C_2 A^1 B^2 + ^3C_3 A^0 B^3$

$(A+B)^3 = 1 \cdot A^3 \cdot 1 + 3 \cdot A^2 \cdot B + 3 \cdot A \cdot B^2 + 1 \cdot 1 \cdot B^3$

$(A+B)^3 = A^3 + 3A^2B + 3AB^2 + B^3$

Now, we substitute $A = 3x^2 + 3a^2$ and $B = -2ax$ back into this expansion and simplify each term.

Term 1: $A^3$

$A^3 = (3x^2 + 3a^2)^3$

This is another binomial expansion of the form $(P+Q)^3$, where $P=3x^2$ and $Q=3a^2$.

$(3x^2 + 3a^2)^3 = ^3C_0 (3x^2)^3 (3a^2)^0 + ^3C_1 (3x^2)^2 (3a^2)^1 + ^3C_2 (3x^2)^1 (3a^2)^2 + ^3C_3 (3x^2)^0 (3a^2)^3$

$= 1 \cdot (27x^6) \cdot 1 + 3 \cdot (9x^4) \cdot (3a^2) + 3 \cdot (3x^2) \cdot (9a^4) + 1 \cdot 1 \cdot (27a^6)$

$= 27x^6 + 81x^4a^2 + 81x^2a^4 + 27a^6$

Term 2: $3A^2B$

$3A^2B = 3 (3x^2 + 3a^2)^2 (-2ax)$

First, expand $(3x^2 + 3a^2)^2$:

$(3x^2 + 3a^2)^2 = (3x^2)^2 + 2(3x^2)(3a^2) + (3a^2)^2 = 9x^4 + 18x^2a^2 + 9a^4$

Now multiply by $3$ and $(-2ax)$:

$3(9x^4 + 18x^2a^2 + 9a^4)(-2ax) = (27x^4 + 54x^2a^2 + 27a^4)(-2ax)$

$= -54ax^5 - 108a^3x^3 - 54a^5x$

Term 3: $3AB^2$

$3AB^2 = 3 (3x^2 + 3a^2) (-2ax)^2$

First, calculate $(-2ax)^2$:

$(-2ax)^2 = (-2)^2 a^2 x^2 = 4a^2x^2$

Now multiply by $3$ and $(3x^2 + 3a^2)$:

$3(3x^2 + 3a^2)(4a^2x^2) = (9x^2 + 9a^2)(4a^2x^2)$

$= 36a^2x^4 + 36a^4x^2$

Term 4: $B^3$

$B^3 = (-2ax)^3$

$= (-2)^3 a^3 x^3 = -8a^3x^3$

Finally, add all the terms together:

$(3x^2 - 2ax + 3a^2)^3 = A^3 + 3A^2B + 3AB^2 + B^3$

$= (27x^6 + 81x^4a^2 + 81x^2a^4 + 27a^6) + (-54ax^5 - 108a^3x^3 - 54a^5x) + (36a^2x^4 + 36a^4x^2) + (-8a^3x^3)$

Combine like terms, grouping by powers of $x$:

$27x^6$

$- 54ax^5$

$+ 81x^4a^2 + 36a^2x^4 = 117a^2x^4$

$- 108a^3x^3 - 8a^3x^3 = -116a^3x^3$

$+ 81x^2a^4 + 36a^4x^2 = 117a^4x^2$

$- 54a^5x$

$+ 27a^6$

So the expansion is:

$27x^6 - 54ax^5 + 117a^2x^4 - 116a^3x^3 + 117a^4x^2 - 54a^5x + 27a^6$