Chapter 8 Binomial Theorem

Given:

The expression to expand is $\left( x^{2} +\frac{3}{x}\right)^{4}$, where $x \neq 0$.

To Find:

The binomial expansion of the given expression.

Solution:

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

$(a+b)^n = \sum_{k=0}^{n} {^nC_k} a^{n-k} b^k$

In this problem, we have $a = x^2$, $b = \frac{3}{x}$, and $n = 4$.

The expansion is:

$\left( x^{2} +\frac{3}{x}\right)^{4} = {^4C_0}(x^2)^4\left(\frac{3}{x}\right)^0 + {^4C_1}(x^2)^3\left(\frac{3}{x}\right)^1 + {^4C_2}(x^2)^2\left(\frac{3}{x}\right)^2 \ $$ + {^4C_3}(x^2)^1\left(\frac{3}{x}\right)^3 + {^4C_4}(x^2)^0\left(\frac{3}{x}\right)^4$

Let's evaluate the binomial coefficients:

$^4C_0 = 1$, $^4C_1 = 4$, $^4C_2 = \frac{4 \times 3}{2} = 6$, $^4C_3 = 4$, $^4C_4 = 1$.

Now, we compute each term of the expansion:

Term 1: $^4C_0(x^2)^4\left(\frac{3}{x}\right)^0 = 1 \cdot x^8 \cdot 1 = x^8$

Term 2: $^4C_1(x^2)^3\left(\frac{3}{x}\right)^1 = 4 \cdot x^6 \cdot \frac{3}{x} = 12x^5$

Term 3: $^4C_2(x^2)^2\left(\frac{3}{x}\right)^2 = 6 \cdot x^4 \cdot \frac{9}{x^2} = 54x^2$

Term 4: $^4C_3(x^2)^1\left(\frac{3}{x}\right)^3 = 4 \cdot x^2 \cdot \frac{27}{x^3} = \frac{108}{x}$

Term 5: $^4C_4(x^2)^0\left(\frac{3}{x}\right)^4 = 1 \cdot 1 \cdot \frac{81}{x^4} = \frac{81}{x^4}$

Combining all the terms, the final expansion is:

$x^8 + 12x^5 + 54x^2 + \frac{108}{x} + \frac{81}{x^4}$

To Compute:

The value of $(98)^5$.

Solution:

We can express 98 as a difference of two numbers that are easy to work with, such as $100 - 2$. Then we can use the Binomial Theorem.

$(98)^5 = (100 - 2)^5$.

Using the Binomial Theorem for $(a-b)^n$ with $a=100$, $b=2$, and $n=5$:

$(a-b)^5 = {^5C_0}a^5 - {^5C_1}a^4b + {^5C_2}a^3b^2 - {^5C_3}a^2b^3 + {^5C_4}ab^4 - {^5C_5}b^5$

The coefficients are: $^5C_0=1, ^5C_1=5, ^5C_2=10, ^5C_3=10, \ $$ ^5C_4=5, ^5C_5=1$.

Substituting the values:

$(100-2)^5 = 1(100)^5 - 5(100)^4(2) + 10(100)^3(2)^2 - 10(100)^2(2)^3 \ $$ + 5(100)^1(2)^4 - 1(100)^0(2)^5$

$= (10^2)^5 - 10 \cdot (10^2)^4 + 10 \cdot (10^2)^3 \cdot 4 - 10 \cdot (10^2)^2 \cdot 8 \ $$ + 5 \cdot 100 \cdot 16 - 1 \cdot 1 \cdot 32$

$= 10^{10} - 10 \cdot 10^8 + 40 \cdot 10^6 - 80 \cdot 10^4 + 8000 - 32$

$= 10,000,000,000 - 1,000,000,000 + 40,000,000 - 800,000 \ $$ + 8,000 - 32$

$= 9,000,000,000 + 40,000,000 - 800,000 + 8,000 - 32$

$= 9,040,000,000 - 800,000 + 7,968$

$= 9,039,200,000 + 7,968$

$= 9,039,207,968$

Therefore, $(98)^5 = 9,039,207,968$.

To Determine:

Which of the two numbers, $(1.01)^{1000000}$ or $10,000$, is larger.

Solution:

We can use the Binomial Theorem to expand $(1.01)^{1000000}$.

First, we write $1.01$ as $(1 + 0.01)$.

So, we need to evaluate $(1 + 0.01)^{1000000}$.

The Binomial Theorem states that for a positive integer $n$:

$(1+x)^n = {^nC_0} + {^nC_1}x + {^nC_2}x^2 + \dots + {^nC_n}x^n$

Here, $x = 0.01$ and $n = 1,000,000$.

Let's look at the first two terms of the expansion:

First term: $^nC_0 = {^{1000000}}C_0 = 1$.

Second term: $^nC_1 x = {^{1000000}}C_1 (0.01) = 1,000,000 \times 0.01 = 10,000$.

The full expansion is:

$(1.01)^{1000000} = 1 + 10,000 + {^{1000000}}C_2(0.01)^2 + \dots$

The third term, ${^{1000000}}C_2(0.01)^2$, and all subsequent terms in the expansion are positive numbers.

So, we can write:

$(1.01)^{1000000} = 10,001 + (\text{sum of other positive terms})$

This shows that the value of $(1.01)^{1000000}$ is greater than $10,001$.

Since $10,001 > 10,000$, it is clear that $(1.01)^{1000000}$ is larger than $10,000$.

Therefore, $(1.01)^{1000000}$ is larger.

To Prove:

$6^n - 5n$ leaves a remainder of 1 when divided by 25 for any positive integer $n$.

This is equivalent to proving that $6^n - 5n - 1$ is a multiple of 25.

Proof:

We start by expressing $6^n$ in a form suitable for the Binomial Theorem. We can write $6 = 1 + 5$.

$6^n = (1+5)^n$

Using the Binomial Theorem, we expand $(1+5)^n$:

$(1+5)^n = {^nC_0}(1)^{n-0}(5)^0 + {^nC_1}(1)^{n-1}(5)^1 + {^nC_2}(1)^{n-2}(5)^2 \ $$ + {^nC_3}(1)^{n-3}(5)^3 + \dots + {^nC_n}(1)^0(5)^n$

$6^n = {^nC_0} + {^nC_1}(5) + {^nC_2}(5^2) + {^nC_3}(5^3) + \dots + {^nC_n}(5^n)$

We know that $^nC_0 = 1$ and $^nC_1 = n$.

$6^n = 1 + n(5) + {^nC_2}(25) + {^nC_3}(125) + \dots + {^nC_n}(5^n)$

$6^n = 1 + 5n + {^nC_2}(25) + {^nC_3}(5 \cdot 25) + \dots$

Now, let's consider the expression $6^n - 5n$:

$6^n - 5n = (1 + 5n + {^nC_2}(25) + {^nC_3}(125) + \dots) - 5n$

$6^n - 5n = 1 + {^nC_2}(25) + {^nC_3}(125) + \dots$

We can factor out 25 from all terms except the first one (for $n \ge 2$).

$6^n - 5n = 1 + 25 \left( {^nC_2} + {^nC_3}(5) + \dots + {^nC_n}5^{n-2} \right)$

Let $K = {^nC_2} + 5 \cdot {^nC_3} + \dots + 5^{n-2} \cdot {^nC_n}$. Since $n$ is a positive integer, all the binomial coefficients are integers, so $K$ is an integer for $n \ge 2$.

If $n=1$, $6^1 - 5(1) = 1$. The expression becomes $1+25(0)$, so $K=0$.

Thus, for any positive integer $n$, we can write:

$6^n - 5n = 1 + 25K$, where $K$ is an integer.

This expression shows that $6^n - 5n$ is 1 more than a multiple of 25. Therefore, when $6^n - 5n$ is divided by 25, it always leaves a remainder of 1.

Hence, proved.

To Expand:

$(1 - 2x)^5$.

Solution:

We use the Binomial Theorem: $(a+b)^n = \sum_{k=0}^{n} {^nC_k} a^{n-k} b^k$.

Here, $a=1$, $b=-2x$, and $n=5$.

The binomial coefficients for $n=5$ are $^5C_0=1, ^5C_1=5, ^5C_2=10, ^5C_3=10, ^5C_4=5, ^5C_5=1$.

The expansion is:

$(1 - 2x)^5 = {^5C_0}(1)^5(-2x)^0 + {^5C_1}(1)^4(-2x)^1 + {^5C_2}(1)^3(-2x)^2 \ $$ + {^5C_3}(1)^2(-2x)^3 + {^5C_4}(1)^1(-2x)^4 + {^5C_5}(1)^0(-2x)^5$

$= 1(1)(1) + 5(1)(-2x) + 10(1)(4x^2) + 10(1)(-8x^3) + 5(1)(16x^4) \ $$ + 1(1)(-32x^5)$

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

Therefore, the expansion is $1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5$.

To Expand:

$\left( \frac{2}{x}-\frac{x}{2} \right)^{5}$.

Solution:

Using the Binomial Theorem with $a=\frac{2}{x}$, $b=-\frac{x}{2}$, and $n=5$.

The coefficients are $1, 5, 10, 10, 5, 1$.

Expansion:

$= {^5C_0}\left(\frac{2}{x}\right)^5 - {^5C_1}\left(\frac{2}{x}\right)^4\left(\frac{x}{2}\right) + {^5C_2}\left(\frac{2}{x}\right)^3\left(\frac{x}{2}\right)^2 - {^5C_3}\left(\frac{2}{x}\right)^2\left(\frac{x}{2}\right)^3 \ $$ + {^5C_4}\left(\frac{2}{x}\right)\left(\frac{x}{2}\right)^4 - {^5C_5}\left(\frac{x}{2}\right)^5$

$= 1\left(\frac{32}{x^5}\right) - 5\left(\frac{16}{x^4}\right)\left(\frac{x}{2}\right) + 10\left(\frac{8}{x^3}\right)\left(\frac{x^2}{4}\right) - 10\left(\frac{4}{x^2}\right)\left(\frac{x^3}{8}\right) \ $$ + 5\left(\frac{2}{x}\right)\left(\frac{x^4}{16}\right) - 1\left(\frac{x^5}{32}\right)$

$= \frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5x^3}{8} - \frac{x^5}{32}$.

Therefore, the expansion is $\frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5x^3}{8} - \frac{x^5}{32}$.

To Expand:

$(2x - 3)^6$.

Solution:

Using the Binomial Theorem with $a=2x$, $b=-3$, and $n=6$.

The coefficients for $n=6$ are $^6C_0=1, ^6C_1=6, ^6C_2=15, \ $$ ^6C_3=20, ^6C_4=15, ^6C_5=6, ^6C_6=1$.

Expansion:

$= {^6C_0}(2x)^6 - {^6C_1}(2x)^5(3) + {^6C_2}(2x)^4(3)^2 - {^6C_3}(2x)^3(3)^3 + \ $$ {^6C_4}(2x)^2(3)^4 - {^6C_5}(2x)(3)^5 + {^6C_6}(3)^6$

$= 1(64x^6) - 6(32x^5)(3) + 15(16x^4)(9) - 20(8x^3)(27) + 15(4x^2)(81) \ $$ - 6(2x)(243) + 1(729)$

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

Therefore, the expansion is $64x^6 - 576x^5 + 2160x^4 \ $$ - 4320x^3 + 4860x^2 - 2916x + 729$.

To Expand:

$\left( \frac{x}{3}+\frac{1}{x} \right)^{5}$.

Solution:

Using the Binomial Theorem with $a=\frac{x}{3}$, $b=\frac{1}{x}$, and $n=5$.

The coefficients are $1, 5, 10, 10, 5, 1$.

Expansion:

$= {^5C_0}\left(\frac{x}{3}\right)^5 + {^5C_1}\left(\frac{x}{3}\right)^4\left(\frac{1}{x}\right) + {^5C_2}\left(\frac{x}{3}\right)^3\left(\frac{1}{x}\right)^2 + {^5C_3}\left(\frac{x}{3}\right)^2\left(\frac{1}{x}\right)^3 \ $$ + {^5C_4}\left(\frac{x}{3}\right)\left(\frac{1}{x}\right)^4 + {^5C_5}\left(\frac{1}{x}\right)^5$

$= 1\left(\frac{x^5}{243}\right) + 5\left(\frac{x^4}{81}\right)\left(\frac{1}{x}\right) + 10\left(\frac{x^3}{27}\right)\left(\frac{1}{x^2}\right) + 10\left(\frac{x^2}{9}\right)\left(\frac{1}{x^3}\right) \ $$ + 5\left(\frac{x}{3}\right)\left(\frac{1}{x^4}\right) + 1\left(\frac{1}{x^5}\right)$

$= \frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}$.

Therefore, the expansion is $\frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}$.

To Expand:

$\left( x+\frac{1}{x} \right)^{6}$.

Solution:

Using the Binomial Theorem with $a=x$, $b=\frac{1}{x}$, and $n=6$.

The coefficients for $n=6$ are $1, 6, 15, 20, 15, 6, 1$.

Expansion:

$= {^6C_0}x^6 + {^6C_1}x^5\left(\frac{1}{x}\right) + {^6C_2}x^4\left(\frac{1}{x}\right)^2 + {^6C_3}x^3\left(\frac{1}{x}\right)^3 + {^6C_4}x^2\left(\frac{1}{x}\right)^4 \ $$ + {^6C_5}x\left(\frac{1}{x}\right)^5 + {^6C_6}\left(\frac{1}{x}\right)^6$

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

$= x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}$.

Therefore, the expansion is $x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}$.

To Evaluate:

$(96)^3$ using the Binomial Theorem.

Solution:

We can write 96 as $(100 - 4)$. So, we need to evaluate $(100 - 4)^3$.

Using the Binomial Theorem for $(a-b)^n$ with $a=100$, $b=4$, and $n=3$:

$(a-b)^3 = {^3C_0}a^3 - {^3C_1}a^2b + {^3C_2}ab^2 - {^3C_3}b^3$

The coefficients are: $^3C_0=1, ^3C_1=3, ^3C_2=3, ^3C_3=1$.

$(100-4)^3 = 1(100)^3 - 3(100)^2(4) + 3(100)(4)^2 - 1(4)^3$

$= 1,000,000 - 3(10,000)(4) + 3(100)(16) - 64$

$= 1,000,000 - 120,000 + 4,800 - 64$

$= 880,000 + 4,736$

$= 884,736$

Therefore, $(96)^3 = 884,736$.

To Evaluate:

$(102)^5$ using the Binomial Theorem.

Solution:

We can write 102 as $(100 + 2)$. So, we need to evaluate $(100 + 2)^5$.

Using the Binomial Theorem for $(a+b)^n$ with $a=100$, $b=2$, and $n=5$:

The coefficients are: $^5C_0=1, ^5C_1=5, ^5C_2=10, \ $$ ^5C_3=10, ^5C_4=5, ^5C_5=1$.

$(100+2)^5 = {^5C_0}(100)^5 + {^5C_1}(100)^4(2) + {^5C_2}(100)^3(2)^2 \ $$ + {^5C_3}(100)^2(2)^3 + {^5C_4}(100)(2)^4 + {^5C_5}(2)^5$

$= 1(10^{10}) + 5(10^8)(2) + 10(10^6)(4) + 10(10^4)(8) + 5(100)(16) + 1(32)$

$= 10,000,000,000 + 1,000,000,000 + 40,000,000 + 800,000 \ $$ + 8,000 + 32$

$= 11,040,808,032$

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

To Evaluate:

$(101)^4$ using the Binomial Theorem.

Solution:

We can write 101 as $(100 + 1)$. So, we need to evaluate $(100 + 1)^4$.

Using the Binomial Theorem with $a=100$, $b=1$, and $n=4$:

The coefficients are: $^4C_0=1, ^4C_1=4, ^4C_2=6, ^4C_3=4, ^4C_4=1$.

$(100+1)^4 = {^4C_0}(100)^4 + {^4C_1}(100)^3 + {^4C_2}(100)^2 + {^4C_3}(100) + {^4C_4}$

$= 1(100,000,000) + 4(1,000,000) + 6(10,000) + 4(100) + 1$

$= 100,000,000 + 4,000,000 + 60,000 + 400 + 1$

$= 104,060,401$

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

To Evaluate:

$(99)^5$ using the Binomial Theorem.

Solution:

We can write 99 as $(100 - 1)$. So, we need to evaluate $(100 - 1)^5$.

Using the Binomial Theorem with $a=100$, $b=1$, and $n=5$:

The coefficients are: $1, 5, 10, 10, 5, 1$.

$(100-1)^5 = {^5C_0}(100)^5 - {^5C_1}(100)^4 + {^5C_2}(100)^3 - {^5C_3}(100)^2 \ $$ + {^5C_4}(100) - {^5C_5}$

$= 1(10^{10}) - 5(10^8) + 10(10^6) - 10(10^4) + 5(100) - 1$

$= 10,000,000,000 - 500,000,000 + 10,000,000 - 100,000 + 500 - 1$

$= 9,500,000,000 + 10,000,000 - 100,000 + 499$

$= 9,510,000,000 - 100,000 + 499$

$= 9,509,900,000 + 499$

$= 9,509,900,499$

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

To Determine:

Which of the two numbers, $(1.1)^{10000}$ or $1000$, is larger.

Solution:

We use the Binomial Theorem to expand $(1.1)^{10000}$.

First, we write $1.1$ as $(1 + 0.1)$.

So, we need to evaluate $(1 + 0.1)^{10000}$.

The Binomial Theorem states that $(1+x)^n = {^nC_0} \ $$ + {^nC_1}x + {^nC_2}x^2 + \dots + {^nC_n}x^n$.

Here, $x = 0.1$ and $n = 10,000$.

The expansion is:

$(1 + 0.1)^{10000} = {^{10000}}C_0 + {^{10000}}C_1(0.1) + {^{10000}}C_2(0.1)^2 + \dots$

Let's evaluate the first two terms:

First term: $^{10000}C_0 = 1$.

Second term: $^{10000}C_1(0.1) = 10,000 \times 0.1 = 1,000$.

The expansion is $(1.1)^{10000} = 1 + 1,000 + (\text{sum of other positive terms})$.

$(1.1)^{10000} = 1,001 + (\text{sum of other positive terms})$.

Since the sum of the other terms is a positive value, we have:

$(1.1)^{10000} > 1,001$.

Since $1,001 > 1,000$, it follows that $(1.1)^{10000}$ is larger than $1,000$.

Therefore, $(1.1)^{10000}$ is larger.

Part 1: Find $(a + b)^4 – (a – b)^4$.

Using the Binomial Theorem:

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

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

Subtracting the second expansion from the first:

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

$= 8a^3b + 8ab^3 = 8ab(a^2 + b^2)$.

Part 2: Evaluate $(\sqrt{3} + \sqrt{2})^4 - (\sqrt{3} - \sqrt{2})^4$.

This expression is in the form $(a+b)^4 - (a-b)^4$ with $a=\sqrt{3}$ and $b=\sqrt{2}$.

Using the result from Part 1, $8ab(a^2 + b^2)$:

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

$= 8(\sqrt{6})(3 + 2)$

$= 8(\sqrt{6})(5) = 40\sqrt{6}$.

Therefore, the value is $40\sqrt{6}$.

Part 1: Find $(x + 1)^6 + (x – 1)^6$.

Using the Binomial Theorem:

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

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

Adding the two expansions, the terms with odd powers cancel out:

$(x+1)^6 + (x-1)^6 = 2(x^6 + 15x^4 + 15x^2 + 1)$.

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

This expression is in the form $(x+1)^6 + (x-1)^6$ with $x=\sqrt{2}$.

Using the result from Part 1, $2(x^6 + 15x^4 + 15x^2 + 1)$:

$= 2((\sqrt{2})^6 + 15(\sqrt{2})^4 + 15(\sqrt{2})^2 + 1)$

$= 2(8 + 15(4) + 15(2) + 1)$

$= 2(8 + 60 + 30 + 1)$

$= 2(99) = 198$.

Therefore, the value is 198.

To Prove:

$9^{n+1} - 8n - 9$ is divisible by 64 for any positive integer $n$.

Proof:

We use the Binomial Theorem to expand $9^{n+1}$. We write $9 = 1 + 8$.

$9^{n+1} = (1+8)^{n+1}$

$(1+8)^{n+1} = {^{n+1}}C_0 + {^{n+1}}C_1(8) + {^{n+1}}C_2(8)^2 + {^{n+1}}C_3(8)^3 + \dots + \ $$ {^{n+1}}C_{n+1}(8)^{n+1}$

$9^{n+1} = 1 + (n+1)(8) + {^{n+1}}C_2(64) + {^{n+1}}C_3(8^3) + \dots$

$9^{n+1} = 1 + 8n + 8 + 64 \cdot {^{n+1}}C_2 + \dots$

$9^{n+1} = 9 + 8n + 64 \left( {^{n+1}}C_2 + 8 \cdot {^{n+1}}C_3 + \dots \right)$

Now, rearrange the equation:

$9^{n+1} - 8n - 9 = 64 \left( {^{n+1}}C_2 + 8 \cdot {^{n+1}}C_3 + \dots \right)$

Let $K = {^{n+1}}C_2 + 8 \cdot {^{n+1}}C_3 + \dots$. Since all terms are integers, $K$ is an integer.

$9^{n+1} - 8n - 9 = 64K$

This shows that $9^{n+1} - 8n - 9$ is a multiple of 64. Therefore, it is divisible by 64.

Hence, proved.

To Prove:

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

Proof:

We start with the Binomial Theorem:

The given summation is $\sum\limits_{r=0}^{n} {^nC_r} \cdot 3^r$.

Comparing this with the general term of the binomial expansion, $^nC_r a^{n-r} b^r$, we can see that if we let $b=3$ and $a=1$, the general term becomes:

$^nC_r (1)^{n-r} (3)^r = {^nC_r} \cdot 1 \cdot 3^r = {^nC_r} \cdot 3^r$.

This matches the term in the given summation.

Therefore, we can substitute $a=1$ and $b=3$ into the Binomial Theorem:

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

$(4)^n = \sum\limits_{r=0}^{n} {^nC_r} \cdot 3^r$

Rearranging the equation, we get:

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

Hence, proved.

Given:

The binomial expansion is $(2 + a)^{50}$.

The 17th term ($T_{17}$) is equal to the 18th term ($T_{18}$).

To Find:

The value of $a$.

Solution:

The general term of the expansion $(x+y)^n$ is $T_{r+1} = {^nC_r} x^{n-r} y^r$.

For the expansion $(2+a)^{50}$, we have $x=2$, $y=a$, and $n=50$.

17th term ($T_{17}$): Here, $r+1 = 17$, so $r=16$.

$T_{17} = {^{50}}C_{16} (2)^{50-16} a^{16} = {^{50}}C_{16} (2)^{34} a^{16}$.

18th term ($T_{18}$): Here, $r+1 = 18$, so $r=17$.

$T_{18} = {^{50}}C_{17} (2)^{50-17} a^{17} = {^{50}}C_{17} (2)^{33} a^{17}$.

Given that $T_{17} = T_{18}$:

${^{50}}C_{16} (2)^{34} a^{16} = {^{50}}C_{17} (2)^{33} a^{17}$.

We can rewrite the equation to solve for $a$. Assuming $a \neq 0$:

$\frac{a^{17}}{a^{16}} = \frac{{^{50}}C_{16} (2)^{34}}{{^{50}}C_{17} (2)^{33}}$

$a = \frac{{^{50}}C_{16}}{{^{50}}C_{17}} \cdot \frac{2^{34}}{2^{33}}$

We use the property $\frac{^nC_k}{^nC_{k+1}} = \frac{k+1}{n-k}$. Here, $n=50$ and $k=16$.

$\frac{{^{50}}C_{16}}{{^{50}}C_{17}} = \frac{16+1}{50-16} = \frac{17}{34} = \frac{1}{2}$.

Also, $\frac{2^{34}}{2^{33}} = 2$.

Substituting these back:

$a = \left(\frac{1}{2}\right) \cdot (2) = 1$.

Therefore, the value of $a=1$.

To Show:

The middle term in the expansion of $(1 + x)^{2n}$ is $\frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{n!} 2n x^n$.

(Note: There appears to be a common typo in this problem statement. The term should be $2^n x^n$, not $2n x^n$. The following proof will derive the correct form.)

Proof:

In the expansion of $(1+x)^{2n}$, the total number of terms is $2n+1$, which is odd. Thus, there is one middle term.

The middle term is the $(\frac{2n}{2} + 1)$-th term, which is the $(n+1)$-th term.

The general term is $T_{r+1} = {^{2n}}C_r x^r$. For the middle term, $r=n$.

Middle term = $T_{n+1} = {^{2n}}C_n x^n$.

Now we evaluate the coefficient ${^{2n}}C_n$:

${^{2n}}C_n = \frac{(2n)!}{n!(2n-n)!} = \frac{(2n)!}{n!n!}$.

We can write $(2n)!$ as the product of all integers from 1 to $2n$.

$(2n)! = [1 \cdot 3 \cdot 5 \dots (2n-1)] \cdot [2 \cdot 4 \cdot 6 \dots (2n)]$.

The second part is the product of all even numbers up to $2n$. We can factor out a 2 from each of the $n$ terms:

$[2 \cdot 4 \cdot 6 \dots (2n)] = 2^n (1 \cdot 2 \cdot 3 \dots n) = 2^n n!$.

So, $(2n)! = [1 \cdot 3 \cdot 5 \dots (2n-1)] \cdot 2^n n!$.

Substituting this back into the coefficient:

${^{2n}}C_n = \frac{[1 \cdot 3 \cdot 5 \dots (2n-1)] \cdot 2^n n!}{n!n!} = \frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{n!} \cdot 2^n$.

Therefore, the middle term is:

$T_{n+1} = \left(\frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{n!} \cdot 2^n\right) x^n = \frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{n!} 2^n x^n$.

This shows that the middle term is $\frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{n!} 2^n x^n$, which confirms the typo in the original problem statement.

Given:

The expansion is $(x + 2y)^9$.

To Find:

The coefficient of the term containing $x^6y^3$.

Solution:

The general term of the expansion $(a+b)^n$ is $T_{r+1} = {^nC_r} a^{n-r} b^r$.

Here, $a=x$, $b=2y$, and $n=9$.

$T_{r+1} = {^9C_r} (x)^{9-r} (2y)^r = {^9C_r} \cdot x^{9-r} \cdot 2^r \cdot y^r = {^9C_r} \cdot 2^r \cdot x^{9-r} y^r$.

We want the term with $x^6y^3$. By comparing the powers, we see that $y^r$ must be $y^3$, so $r=3$.

Let's check the power of $x$: $x^{9-r} = x^{9-3} = x^6$. This matches.

So, we need to find the coefficient of the term where $r=3$.

The coefficient is ${^9C_3} \cdot 2^3$.

$^9C_3 = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84$.

$2^3 = 8$.

Coefficient = $84 \times 8 = 672$.

Therefore, the coefficient of $x^6y^3$ is 672.

Given:

For the expansion of $(x+a)^n$:

Second term, $T_2 = {^nC_1} x^{n-1} a = nx^{n-1}a = 240$. ...(i)

Third term, $T_3 = {^nC_2} x^{n-2} a^2 = \frac{n(n-1)}{2}x^{n-2}a^2 = 720$. ...(ii)

Fourth term, $T_4 = {^nC_3} x^{n-3} a^3 = \frac{n(n-1)(n-2)}{6}x^{n-3}a^3 = 1080$. ...(iii)

Solution:

Step 1: Find n.

Divide equation (ii) by (i): $\frac{T_3}{T_2} = \frac{\frac{n(n-1)}{2}x^{n-2}a^2}{nx^{n-1}a} = \frac{720}{240} = 3$.

$\frac{n-1}{2} \cdot \frac{a}{x} = 3 \implies (n-1)\frac{a}{x} = 6$. ...(iv)

Divide equation (iii) by (ii): $\frac{T_4}{T_3} = \frac{\frac{n(n-1)(n-2)}{6}x^{n-3}a^3}{\frac{n(n-1)}{2}x^{n-2}a^2} = \frac{1080}{720} = \frac{3}{2}$.

$\frac{n-2}{3} \cdot \frac{a}{x} = \frac{3}{2} \implies (n-2)\frac{a}{x} = \frac{9}{2}$. ...(v)

Now divide (iv) by (v): $\frac{(n-1)\frac{a}{x}}{(n-2)\frac{a}{x}} = \frac{6}{9/2} = 6 \cdot \frac{2}{9} = \frac{4}{3}$.

$\frac{n-1}{n-2} = \frac{4}{3} \implies 3(n-1) = 4(n-2) \implies 3n-3 = 4n-8 \implies n=5$.

Step 2: Find a and x.

Substitute $n=5$ into (iv): $(5-1)\frac{a}{x} = 6 \implies 4\frac{a}{x} = 6 \implies \frac{a}{x} = \frac{3}{2}$. So, $a = \frac{3}{2}x$.

Substitute $n=5$ and $a = \frac{3}{2}x$ into (i):

$5x^{5-1}(\frac{3}{2}x) = 240 \implies 5x^4(\frac{3}{2}x) = 240 \implies \frac{15}{2}x^5 = 240$.

$x^5 = 240 \cdot \frac{2}{15} = 16 \cdot 2 = 32$.

Since $x^5 = 32$, we get $x=2$.

Now find $a$: $a = \frac{3}{2}x = \frac{3}{2}(2) = 3$.

Therefore, $x=2, a=3, n=5$.

Given:

In the binomial expansion of $(1 + a)^n$, the coefficients of three consecutive terms are in the ratio $1:7:42$.

To Find:

The value of $n$.

Solution:

The coefficient of the $(r+1)^{\text{th}}$ term in the expansion of $(1+a)^n$ is $^nC_r$.

Let the three consecutive terms be the $(r+1)^{\text{th}}$, $(r+2)^{\text{th}}$, and $(r+3)^{\text{th}}$ terms.

Their respective coefficients are $^nC_r$, $^nC_{r+1}$, and $^nC_{r+2}$.

We are given the ratio:

$^nC_r : {^nC_{r+1}} : {^nC_{r+2}} = 1 : 7 : 42$

From this ratio, we can form two separate equations:

We use the property relating consecutive binomial coefficients: $\frac{^nC_{k+1}}{^nC_k} = \frac{n-k}{k+1}$.

From equation (i), we can write $\frac{^nC_{r+1}}{^nC_r} = 7$.

Applying the property with $k=r$:

$\frac{n-r}{r+1} = 7$

$n-r = 7(r+1)$

$n-r = 7r + 7$

$n - 8r = 7$

From equation (ii), we can write $\frac{^nC_{r+2}}{^nC_{r+1}} = 6$.

Applying the property with $k=r+1$:

$\frac{n-(r+1)}{r+1+1} = 6$

$\frac{n-r-1}{r+2} = 6$

$n-r-1 = 6(r+2)$

$n-r-1 = 6r + 12$

$n - 7r = 13$

Now we have a system of two linear equations with variables $n$ and $r$.

From (iii), $n = 8r + 7$.

Substitute this expression for $n$ into equation (iv):

$(8r + 7) - 7r - 13 = 0$

$r - 6 = 0$

$r = 6$

Now, substitute the value of $r=6$ back into the expression for $n$:

$n = 8(6) + 7$

$n = 48 + 7$

$n = 55$

Therefore, the value of $n=55$.

Given:

The binomial expression is $(x + 3)^8$.

To Find:

The coefficient of the term containing $x^5$.

Solution:

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

For $(x+3)^8$, we have $a=x$, $b=3$, and $n=8$.

$T_{r+1} = {^8C_r} (x)^{8-r} (3)^r = {^8C_r} \cdot 3^r \cdot x^{8-r}$.

We want the term where the power of $x$ is 5. So, we set $8-r = 5$, which gives $r=3$.

The term is $T_{3+1} = T_4$. The coefficient is ${^8C_3} \cdot 3^3$.

$^8C_3 = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.

$3^3 = 27$.

Coefficient = $56 \times 27 = 1512$.

Therefore, the coefficient of $x^5$ is 1512.

Given:

The binomial expression is $(a - 2b)^{12}$.

To Find:

The coefficient of the term containing $a^5b^7$.

Solution:

The general term is $T_{r+1} = {^{12}C_r} (a)^{12-r} (-2b)^r = {^{12}C_r} \cdot (-2)^r \cdot a^{12-r} \cdot b^r$.

We want the term with $a^5b^7$. By comparing the powers of $b$, we get $r=7$.

Let's check the power of $a$: $12-r = 12-7=5$. This matches.

The coefficient is ${^{12}}C_7 \cdot (-2)^7$.

$^{12}C_7 = {^{12}}C_5 = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792$.

$(-2)^7 = -128$.

Coefficient = $792 \times (-128) = -101,376$.

Therefore, the coefficient of $a^5b^7$ is -101,376.

Given:

The binomial expression is $(x^2 - y)^6$.

To Find:

The general term in the expansion of $(x^2 - y)^6$.

Solution:

The general term, which is the $(r+1)^{\text{th}}$ term in the binomial expansion of $(a+b)^n$, is given by the formula:

For the given expression $(x^2 - y)^6$, we identify the components:

$a = x^2$

$b = -y$

$n = 6$

Now, we substitute these values into the general term formula:

$T_{r+1} = {^6C_r} (x^2)^{6-r} (-y)^r$

We can simplify this expression by applying the rules of exponents and separating the negative sign:

$T_{r+1} = {^6C_r} x^{2(6-r)} (-1)^r y^r$

$T_{r+1} = (-1)^r {^6C_r} x^{12-2r} y^r$

Here, the value of $r$ can range from 0 to 6, i.e., $r \in \{0, 1, 2, 3, 4, 5, 6\}$.

Therefore, the general term in the expansion of $(x^2 - y)^6$ is $T_{r+1} = (-1)^r {^6C_r} x^{12-2r} y^r$.

Given:

The binomial expression is $(x^2 - yx)^{12}$, with the condition $x \neq 0$.

To Find:

The general term in the expansion of $(x^2 - yx)^{12}$.

Solution:

The formula for the general term, the $(r+1)^{\text{th}}$ term, in the binomial expansion of $(a+b)^n$ is:

For the given expression $(x^2 - yx)^{12}$, we identify:

$a = x^2$

$b = -yx$

$n = 12$

Substitute these values into the general term formula:

$T_{r+1} = {^{12}C_r} (x^2)^{12-r} (-yx)^r$

Now, we simplify the expression using the properties of exponents:

$T_{r+1} = {^{12}C_r} (x^{2(12-r)}) ((-1)^r y^r x^r)$

$T_{r+1} = {^{12}C_r} (x^{24-2r}) (-1)^r y^r x^r$

Combine the terms with the same base, $x$:

$T_{r+1} = (-1)^r {^{12}C_r} x^{24-2r+r} y^r$

$T_{r+1} = (-1)^r {^{12}C_r} x^{24-r} y^r$

Here, the value of $r$ can range from 0 to 12, i.e., $r \in \{0, 1, 2, \dots, 12\}$.

Therefore, the general term in the expansion of $(x^2 - yx)^{12}$ is $T_{r+1} = (-1)^r {^{12}C_r} x^{24-r} y^r$.

Given:

The binomial expansion is $(x - 2y)^{12}$.

To Find:

The 4th term of the expansion.

Solution:

The general term of the expansion $(a+b)^n$ is $T_{r+1} = {^nC_r} a^{n-r} b^r$.

For $(x - 2y)^{12}$, we have $a=x$, $b=-2y$, and $n=12$.

To find the 4th term, we need to set $r+1=4$, which means $r=3$.

Substitute these values into the formula:

$T_4 = T_{3+1} = {^{12}C_3} (x)^{12-3} (-2y)^3$

$T_4 = {^{12}C_3} x^9 (-2)^3 y^3$

Now, we calculate the coefficient:

$^{12}C_3 = \frac{12!}{3!(12-3)!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 2 \times 11 \times 10 = 220$.

$(-2)^3 = -8$.

The coefficient is $220 \times (-8) = -1760$.

So, the 4th term is $-1760x^9y^3$.

Therefore, the 4th term is $-1760x^9y^3$.

Given:

The expansion is $\left( 9x-\frac{1}{3\sqrt{x}} \right)^{18}$.

To Find:

The 13th term of the expansion.

Solution:

The general term is $T_{r+1} = {^nC_r} a^{n-r} b^r$.

Here, $a=9x$, $b=-\frac{1}{3\sqrt{x}}$, and $n=18$.

To find the 13th term, we set $r+1=13$, which means $r=12$.

$T_{13} = T_{12+1} = {^{18}}C_{12} (9x)^{18-12} \left(-\frac{1}{3\sqrt{x}}\right)^{12}$

$T_{13} = {^{18}}C_{12} (9x)^6 \left(\frac{1}{3\sqrt{x}}\right)^{12}$ (since the power 12 is even, the negative sign becomes positive)

$T_{13} = {^{18}}C_{12} (9^6 x^6) \left(\frac{1}{3^{12} (\sqrt{x})^{12}}\right)$

Since $9=3^2$, $9^6=(3^2)^6=3^{12}$. Also, $(\sqrt{x})^{12} = (x^{1/2})^{12} = x^6$.

$T_{13} = {^{18}}C_{12} (3^{12} x^6) \left(\frac{1}{3^{12} x^6}\right)$

The terms involving variables and numbers cancel out: $T_{13} = {^{18}}C_{12}$.

Now, we calculate the value of $^{18}C_{12}$:

$^{18}C_{12} = {^{18}}C_{18-12} = {^{18}}C_6 = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 18,564$.

Therefore, the 13th term is 18,564.

Given:

The expansion is $\left( 3 - \frac{x^3}{6} \right)^7$.

To Find:

The middle terms of the expansion.

Solution:

Here, the power is $n=7$. The total number of terms is $n+1 = 8$, which is even. So, there are two middle terms.

The positions of the middle terms are $(\frac{n+1}{2})$-th and $(\frac{n+1}{2}+1)$-th.

Middle terms are the $(\frac{7+1}{2})$-th = 4th term, and the $(\frac{7+1}{2}+1)$-th = 5th term.

4th Term ($T_4$): Here $r=3$.

$T_4 = {^7C_3} (3)^{7-3} \left(-\frac{x^3}{6}\right)^3 = 35 \cdot 3^4 \cdot (-\frac{x^9}{6^3}) = 35 \cdot 81 \cdot (-\frac{x^9}{216}) \ $$ = -\frac{2835}{216}x^9 = -\frac{105}{8}x^9$.

5th Term ($T_5$): Here $r=4$.

$T_5 = {^7C_4} (3)^{7-4} \left(-\frac{x^3}{6}\right)^4 = 35 \cdot 3^3 \cdot (\frac{x^{12}}{6^4}) = 35 \cdot 27 \cdot (\frac{x^{12}}{1296}) \ $$ = \frac{945}{1296}x^{12} = \frac{35}{48}x^{12}$.

The middle terms are $-\frac{105}{8}x^9$ and $\frac{35}{48}x^{12}$.

Given:

The expansion is $\left( \frac{x}{3}+9y \right)^{10}$.

To Find:

The middle term of the expansion.

Solution:

Here, the power is $n=10$. The total number of terms is $n+1=11$, which is odd. So, there is one middle term.

The position of the middle term is $(\frac{n}{2}+1)$-th = $(\frac{10}{2}+1)$-th = 6th term.

6th Term ($T_6$): Here $r=5$.

$T_6 = {^{10}}C_5 \left(\frac{x}{3}\right)^{10-5} (9y)^5 = {^{10}}C_5 \left(\frac{x^5}{3^5}\right) (9^5y^5)$.

Since $9=3^2$, $9^5 = (3^2)^5=3^{10}$.

$T_6 = {^{10}}C_5 \frac{x^5}{3^5} \cdot 3^{10}y^5 = {^{10}}C_5 \cdot 3^{10-5} x^5 y^5 = {^{10}}C_5 \cdot 3^5 x^5 y^5$.

$^{10}C_5 = \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 252$.

$3^5 = 243$.

Coefficient = $252 \times 243 = 61,236$.

The middle term is $61,236 x^5 y^5$.

To Prove:

In the expansion of $(1+a)^{m+n}$, the coefficient of $a^m$ is equal to the coefficient of $a^n$.

Proof:

The general term in the binomial expansion of $(1+x)^N$ is $T_{r+1} = {^NC_r} x^r$.

For the expansion of $(1+a)^{m+n}$, we have $x=a$ and the power $N=m+n$.

The general term is $T_{r+1} = {^{m+n}}C_r a^r$.

The coefficient of the term $a^r$ is ${^{m+n}}C_r$.

Coefficient of $a^m$:

To find the coefficient of $a^m$, we set $r=m$.

Coefficient of $a^m = {^{m+n}}C_m$.

Using the combination formula, ${^NC_k} = \frac{N!}{k!(N-k)!}$:

${^{m+n}}C_m = \frac{(m+n)!}{m!((m+n)-m)!} = \frac{(m+n)!}{m!n!}$.

Coefficient of $a^n$:

To find the coefficient of $a^n$, we set $r=n$.

Coefficient of $a^n = {^{m+n}}C_n$.

Using the combination formula:

${^{m+n}}C_n = \frac{(m+n)!}{n!((m+n)-n)!} = \frac{(m+n)!}{n!m!}$.

Comparison:

We see that the coefficient of $a^m$ is $\frac{(m+n)!}{m!n!}$ and the coefficient of $a^n$ is $\frac{(m+n)!}{n!m!}$.

Since multiplication is commutative, $m!n! = n!m!$.

Therefore, ${^{m+n}}C_m = {^{m+n}}C_n$.

This is also a direct application of the property ${^NC_k} = {^NC_{N-k}}$, where $N=m+n$ and $k=m$, so $N-k = (m+n)-m = n$.

Hence, the coefficients of $a^m$ and $a^n$ are equal. Proved.

Given:

In the expansion of $(x + 1)^n$, the coefficients of the $(r-1)^{\text{th}}$, $r^{\text{th}}$, and $(r+1)^{\text{th}}$ terms are in the ratio $1:3:5$.

Solution:

The coefficient of the $k^{\text{th}}$ term in the expansion of $(x+1)^n$ is $^nC_{k-1}$.

Coefficient of $(r-1)^{\text{th}}$ term = $^nC_{r-2}$.

Coefficient of $r^{\text{th}}$ term = $^nC_{r-1}$.

Coefficient of $(r+1)^{\text{th}}$ term = $^nC_r$.

We are given the ratios:

1) $\frac{^nC_{r-2}}{^nC_{r-1}} = \frac{1}{3} \implies \frac{r-1}{n-(r-1)+1} = \frac{r-1}{n-r+2} = \frac{1}{3} \implies 3r-3 = n-r+2 \ $$ \implies n - 4r + 5 = 0$. ...(i)

2) $\frac{^nC_{r-1}}{^nC_{r}} = \frac{3}{5} \implies \frac{r}{n-r+1} = \frac{3}{5} \implies 5r = 3(n-r+1) \ $$ \implies 5r = 3n-3r+3 \ $$ \implies 3n - 8r + 3 = 0$. ...(ii)

From (i), $n = 4r - 5$. Substitute this into (ii):

$3(4r - 5) - 8r + 3 = 0$

$12r - 15 - 8r + 3 = 0$

$4r - 12 = 0 \implies 4r = 12 \implies r=3$.

Now, find $n$ using $n = 4r - 5$:

$n = 4(3) - 5 = 12 - 5 = 7$.

Therefore, $n=7$ and $r=3$.

To Prove:

Coefficient of $x^n$ in $(1 + x)^{2n} = 2 \times ($Coefficient of $x^n$ in $(1 + x)^{2n-1})$.

Proof:

The coefficient of $x^k$ in the expansion of $(1+x)^N$ is $^NC_k$.

Step 1: Find the coefficient of $x^n$ in $(1 + x)^{2n}$.

Here, $N=2n$ and $k=n$. The coefficient is ${^{2n}}C_n$.

${^{2n}}C_n = \frac{(2n)!}{n!(2n-n)!} = \frac{(2n)!}{n!n!}$.

Step 2: Find the coefficient of $x^n$ in $(1 + x)^{2n-1}$.

Here, $N=2n-1$ and $k=n$. The coefficient is ${^{2n-1}}C_n$.

${^{2n-1}}C_n = \frac{(2n-1)!}{n!((2n-1)-n)!} = \frac{(2n-1)!}{n!(n-1)!}$.

Step 3: Show the relationship.

We need to prove that ${^{2n}}C_n = 2 \cdot {^{2n-1}}C_n$.

Let's start with the expression for ${^{2n}}C_n$ and manipulate it:

${^{2n}}C_n = \frac{(2n)!}{n!n!} = \frac{2n \cdot (2n-1)!}{n \cdot (n-1)! \cdot n!} = \frac{2n}{n} \cdot \frac{(2n-1)!}{n!(n-1)!}$

${^{2n}}C_n = 2 \cdot \frac{(2n-1)!}{n!(n-1)!}$

The term $\frac{(2n-1)!}{n!(n-1)!}$ is exactly the expression for ${^{2n-1}}C_n$.

Therefore, ${^{2n}}C_n = 2 \cdot {^{2n-1}}C_n$.

Hence, proved.

Given:

In the expansion of $(1 + x)^m$, the coefficient of $x^2$ is 6.

To Find:

A positive value of $m$.

Solution:

The general term in the expansion of $(1+x)^m$ is $T_{r+1} = {^mC_r} x^r$.

The coefficient of $x^r$ is $^mC_r$.

For the term with $x^2$, we have $r=2$.

The coefficient of $x^2$ is $^mC_2$.

We are given that this coefficient is 6.

$^mC_2 = 6$

Using the combination formula:

$\frac{m!}{2!(m-2)!} = 6$

$\frac{m(m-1)}{2 \times 1} = 6$

$m(m-1) = 12$

$m^2 - m - 12 = 0$

Factoring the quadratic equation:

$(m-4)(m+3) = 0$

The possible values for $m$ are $m=4$ or $m=-3$.

Since we are looking for a positive value of $m$, the solution is $m=4$.

Given:

The expansion is $\left( \frac{3}{2}x^2 - \frac{1}{3x} \right)^6$.

To Find:

The term independent of $x$.

Solution:

The general term is $T_{r+1} = {^nC_r} a^{n-r} b^r$.

Here, $a=\frac{3}{2}x^2$, $b=-\frac{1}{3x}$, and $n=6$.

$T_{r+1} = {^6C_r} \left(\frac{3}{2}x^2\right)^{6-r} \left(-\frac{1}{3x}\right)^r$

$T_{r+1} = {^6C_r} \left(\frac{3}{2}\right)^{6-r} (x^2)^{6-r} (-1)^r \left(\frac{1}{3}\right)^r \left(\frac{1}{x}\right)^r$

$T_{r+1} = {^6C_r} (-1)^r \frac{3^{6-r}}{2^{6-r}} \frac{1}{3^r} x^{12-2r} x^{-r}$

$T_{r+1} = {^6C_r} (-1)^r \frac{3^{6-2r}}{2^{6-r}} x^{12-3r}$.

For the term to be independent of $x$, the power of $x$ must be 0.

$12-3r=0 \implies 3r=12 \implies r=4$.

The term is the 5th term ($T_5$). We substitute $r=4$ into the coefficient part:

Term = ${^6C_4} (-1)^4 \frac{3^{6-2(4)}}{2^{6-4}} = {^6C_2} \cdot 1 \cdot \frac{3^{-2}}{2^2} = 15 \cdot \frac{1}{3^2 \cdot 2^2} \ $$ = 15 \cdot \frac{1}{9 \cdot 4} \ $$ = \frac{15}{36} = \frac{5}{12}$.

The term independent of $x$ is $\frac{5}{12}$.

Given:

In the expansion of $(1+a)^n$, the coefficients of $a^{r-1}$, $a^r$, and $a^{r+1}$ are in Arithmetic Progression (AP).

The coefficients are ${^nC_{r-1}}, {^nC_r}, {^nC_{r+1}}$.

To Prove:

$n^2 - n(4r+1) + 4r^2 - 2 = 0$.

Proof:

If three terms are in AP, then the middle term is the average of the other two. So, $2 \cdot {^nC_r} = {^nC_{r-1}} + {^nC_{r+1}}$.

Divide by ${^nC_r}$: $2 = \frac{^nC_{r-1}}{^nC_r} + \frac{^nC_{r+1}}{^nC_r}$.

Using the property $\frac{^nC_{k-1}}{^nC_k} = \frac{k}{n-k+1}$, we get:

$2 = \frac{r}{n-r+1} + \frac{n-r}{r+1}$.

$2(n-r+1)(r+1) = r(r+1) + (n-r)(n-r+1)$.

$2(nr+n-r^2-r+r+1) = r^2+r + n^2-nr+n-nr+r^2-r$.

$2(nr+n-r^2+1) = 2r^2 + n^2 - 2nr + n$.

$2nr+2n-2r^2+2 = 2r^2 + n^2 - 2nr + n$.

Rearranging all terms to one side:

$n^2 - 4nr - n + 4r^2 - 2 = 0$.

$n^2 - n(4r+1) + 4r^2 - 2 = 0$.

Hence, proved.

To Show:

Coefficient of middle term in $(1+x)^{2n}$ equals the sum of coefficients of the two middle terms in $(1+x)^{2n-1}$.

Proof:

1. Middle term of $(1+x)^{2n}$:

The power is $2n$ (even), so there are $2n+1$ (odd) terms. The middle term is the $(\frac{2n}{2}+1) = (n+1)$-th term.

The coefficient of the $(n+1)$-th term ($T_{n+1}$) is ${^{2n}}C_n$.

2. Middle terms of $(1+x)^{2n-1}$:

The power is $2n-1$ (odd), so there are $2n$ (even) terms. There are two middle terms:

The $(\frac{2n-1+1}{2}) = n$-th term and the $(n+1)$-th term.

Coefficient of the $n$-th term ($T_n$) is ${^{2n-1}}C_{n-1}$.

Coefficient of the $(n+1)$-th term ($T_{n+1}$) is ${^{2n-1}}C_n$.

3. Show the equality:

We need to prove that ${^{2n}}C_n = {^{2n-1}}C_{n-1} + {^{2n-1}}C_n$.

This is a direct application of Pascal's Identity, which states that ${^N}C_k = {^{N-1}}C_{k-1} + {^{N-1}}C_k$.

By letting $N=2n$ and $k=n$, we get the required identity.

Hence, proved.

To Find:

The coefficient of $a^4$ in the product $(1 + 2a)^4 (2 – a)^5$.

Solution:

First, we write out the expansions for both terms:

$(1+2a)^4 = {^4C_0}(2a)^0 + {^4C_1}(2a)^1 + {^4C_2}(2a)^2 + {^4C_3}(2a)^3 + {^4C_4}(2a)^4 \ $$ = 1 + 8a + 24a^2 + 32a^3 + 16a^4$.

$(2-a)^5 = {^5C_0}2^5 - {^5C_1}2^4a + {^5C_2}2^3a^2 - {^5C_3}2^2a^3 + {^5C_4}2a^4 - {^5C_5}a^5 \ $$ = 32 - 80a + 80a^2 - 40a^3 + 10a^4 - a^5$.

To get the term with $a^4$ in the product, we multiply terms from each expansion whose powers of 'a' sum to 4:

• (Constant from 1st) $\times$ ($a^4$ from 2nd): $(1)(10a^4) = 10a^4$. Coeff: 10.
• ($a^1$ from 1st) $\times$ ($a^3$ from 2nd): $(8a)(-40a^3) = -320a^4$. Coeff: -320.
• ($a^2$ from 1st) $\times$ ($a^2$ from 2nd): $(24a^2)(80a^2) = 1920a^4$. Coeff: 1920.
• ($a^3$ from 1st) $\times$ ($a^1$ from 2nd): $(32a^3)(-80a) = -2560a^4$. Coeff: -2560.
• ($a^4$ from 1st) $\times$ (Constant from 2nd): $(16a^4)(32) = 512a^4$. Coeff: 512.

The total coefficient of $a^4$ is the sum of these individual coefficients:

$10 - 320 + 1920 - 2560 + 512 = -438$.

The coefficient is -438.

To Find:

The $r^{\text{th}}$ term from the end in the expansion of $(x + a)^n$.

Solution:

The expansion of $(x+a)^n$ has $n+1$ terms in total.

The $r^{\text{th}}$ term from the end is the same as the $( (n+1) - r + 1 )^{\text{th}}$ term from the beginning.

Position from the beginning = $(n-r+2)^{\text{th}}$ term.

The general term from the beginning is $T_{k+1} = {^nC_k} x^{n-k} a^k$.

For the $(n-r+2)^{\text{th}}$ term, we set $k+1 = n-r+2$, which means $k = n-r+1$.

The term is: $T_{n-r+2} = {^nC_{n-r+1}} x^{n-(n-r+1)} a^{n-r+1}$.

$= {^nC_{n-r+1}} x^{r-1} a^{n-r+1}$.

Using the property ${^nC_k} = {^nC_{n-k}}$, we have ${^nC_{n-r+1}} = {^nC_{n-(n-r+1)}} = {^nC_{r-1}}$.

So, the term is ${^nC_{r-1}} x^{r-1} a^{n-r+1}$.

Alternate Method:

The $r^{\text{th}}$ term from the end of $(x+a)^n$ is the same as the $r^{\text{th}}$ term from the beginning of $(a+x)^n$.

For $(a+x)^n$, the $r^{\text{th}}$ term is $T_r = {^nC_{r-1}} a^{n-(r-1)} x^{r-1} = {^nC_{r-1}} a^{n-r+1} x^{r-1}$.

Both methods give the same result. The term is ${^nC_{r-1}} x^{r-1} a^{n-r+1}$.

Given:

The expansion is $\left( x^{1/3} + \frac{1}{2x^{1/3}} \right)^{18}$.

To Find:

The term independent of $x$.

Solution:

The general term is $T_{r+1} = {^{18}C_r} (x^{1/3})^{18-r} \left(\frac{1}{2x^{1/3}}\right)^r$.

$T_{r+1} = {^{18}C_r} x^{\frac{18-r}{3}} \cdot \left(\frac{1}{2}\right)^r \cdot (x^{-1/3})^r \ $$ = {^{18}C_r} \left(\frac{1}{2}\right)^r x^{\frac{18-r}{3} - \frac{r}{3}} = {^{18}C_r} \left(\frac{1}{2}\right)^r x^{\frac{18-2r}{3}}$.

For the term to be independent of $x$, the power of $x$ must be 0.

$\frac{18-2r}{3} = 0 \implies 18-2r=0 \implies 2r=18 \implies r=9$.

The term is the 10th term ($T_{10}$). We substitute $r=9$ into the coefficient part:

Term = ${^{18}}C_9 \left(\frac{1}{2}\right)^9$.

$^{18}C_9 = \frac{18!}{9!9!} = 48,620$.

Term = $48,620 \cdot \frac{1}{512} = \frac{48620}{512} = \frac{12155}{128}$.

The term is $\frac{12155}{128}$.

Given:

The sum of coefficients of the first three terms of $\left( x-\frac{3}{x^{2}} \right)^{m}$ is 559.

Solution:

The first three terms are:

$T_1 = {^mC_0} x^m = x^m$. Coefficient = 1.

$T_2 = {^mC_1} x^{m-1} (-\frac{3}{x^2}) = -3m x^{m-3}$. Coefficient = $-3m$.

$T_3 = {^mC_2} x^{m-2} (-\frac{3}{x^2})^2 = \frac{m(m-1)}{2} x^{m-2} (\frac{9}{x^4}) = \frac{9m(m-1)}{2} x^{m-6}$. Coefficient = $\frac{9m(m-1)}{2}$.

Sum of coefficients: $1 - 3m + \frac{9m(m-1)}{2} = 559$.

$2 - 6m + 9m^2 - 9m = 1118 \implies 9m^2 - 15m - 1116 = 0$.

Divide by 3: $3m^2 - 5m - 372 = 0$.

Using the quadratic formula, $m = \frac{5 \pm \sqrt{25 - 4(3)(-372)}}{6} = \frac{5 \pm \sqrt{4489}}{6} = \frac{5 \pm 67}{6}$.

Since $m$ is a natural number, $m = \frac{5+67}{6} = \frac{72}{6} = 12$.

Now, find the term with $x^3$ in $\left( x-\frac{3}{x^{2}} \right)^{12}$.

The general term's power of $x$ is $x^{m-3r} = x^{12-3r}$.

$12-3r = 3 \implies 3r=9 \implies r=3$.

The term is $T_4 = {^{12}}C_3 x^{12-3} (-\frac{3}{x^2})^3 = 220 \cdot x^9 \cdot (-\frac{27}{x^6}) = -5940x^3$.

The term is $-5940x^3$.

Given:

In the expansion of $(1+x)^{34}$, the coefficients of the $(r-5)^{\text{th}}$ and $(2r-1)^{\text{th}}$ terms are equal.

Solution:

The coefficient of the $k^{\text{th}}$ term in $(1+x)^n$ is $^nC_{k-1}$.

Coefficient of $(r-5)^{\text{th}}$ term = $^{34}C_{r-6}$.

Coefficient of $(2r-1)^{\text{th}}$ term = $^{34}C_{2r-2}$.

Given they are equal: $^{34}C_{r-6} = {^{34}}C_{2r-2}$.

Using the property $^nC_a = ^nC_b \implies a=b$ or $a+b=n$.

Case 1: $r-6 = 2r-2 \implies r = -4$. This is not possible as term numbers must be positive.

Case 2: $(r-6) + (2r-2) = 34 \implies 3r - 8 = 34 \implies 3r = 42 \ $$ \implies r = 14$.

For $r=14$, the terms are the 9th and 27th. The coefficients are $^{34}C_8$ and $^{34}C_{26}$. Since $^{34}C_8 = {^{34}}C_{34-8} = {^{34}}C_{26}$, this solution is valid.

Therefore, $r=14$.

Given:

In the expansion of $(a+b)^n$:

First term, $T_1 = {^nC_0}a^n = a^n = 729$. ...(i)

Second term, $T_2 = {^nC_1}a^{n-1}b = na^{n-1}b = 7290$. ...(ii)

Third term, $T_3 = {^nC_2}a^{n-2}b^2 = \frac{n(n-1)}{2}a^{n-2}b^2 = 30375$. ...(iii)

Solution:

Step 1: Find n.

Divide equation (ii) by (i): $\frac{na^{n-1}b}{a^n} = \frac{7290}{729} \implies n\frac{b}{a} = 10$. ...(iv)

Divide equation (iii) by (ii): $\frac{\frac{n(n-1)}{2}a^{n-2}b^2}{na^{n-1}b} = \frac{30375}{7290} = \frac{25}{6}$.

$\frac{n-1}{2} \cdot \frac{b}{a} = \frac{25}{6}$. ...(v)

From (iv), $\frac{b}{a} = \frac{10}{n}$. Substitute this into (v):

$\frac{n-1}{2} \cdot \frac{10}{n} = \frac{25}{6} \implies \frac{5(n-1)}{n} = \frac{25}{6} \implies \frac{n-1}{n} = \frac{5}{6}$.

$6(n-1)=5n \implies 6n-6=5n \implies n=6$.

Step 2: Find a and b.

Substitute $n=6$ into (iv): $6\frac{b}{a} = 10 \implies \frac{b}{a} = \frac{10}{6} = \frac{5}{3}$. So, $b=\frac{5}{3}a$.

Substitute $n=6$ into (i): $a^6 = 729$. Since $3^6 = 729$, we get $a=3$.

Now find $b$: $b = \frac{5}{3}a = \frac{5}{3}(3) = 5$.

Therefore, $a=3, b=5, n=6$.

Given:

In the expansion of $(3 + ax)^9$, the coefficients of $x^2$ and $x^3$ are equal.

Solution:

The general term in the expansion of $(3+ax)^9$ is $T_{r+1} = {^9C_r} (3)^{9-r} (ax)^r = {^9C_r} 3^{9-r} a^r x^r$.

The coefficient of $x^r$ is ${^9C_r} 3^{9-r} a^r$.

Coefficient of $x^2$ (for r=2):

Coefficient = ${^9C_2} 3^{9-2} a^2 = {^9C_2} 3^7 a^2$.

Coefficient of $x^3$ (for r=3):

Coefficient = ${^9C_3} 3^{9-3} a^3 = {^9C_3} 3^6 a^3$.

Given that the coefficients are equal:

${^9C_2} 3^7 a^2 = {^9C_3} 3^6 a^3$.

Assuming $a \neq 0$, we can divide by $3^6 a^2$:

${^9C_2} \cdot 3 = {^9C_3} \cdot a$.

$a = \frac{3 \cdot {^9C_2}}{^9C_3} = 3 \cdot \frac{9!/(2!7!)}{9!/(3!6!)} = 3 \cdot \frac{3!6!}{2!7!} = 3 \cdot \frac{3 \cdot 2! \cdot 6!}{2! \cdot 7 \cdot 6!} = 3 \cdot \frac{3}{7} = \frac{9}{7}$.

Therefore, $a = \frac{9}{7}$.

To Find:

The coefficient of $x^5$ in the product $(1 + 2x)^6 (1 – x)^7$.

Solution:

First, we write out the expansions for both terms:

$(1+2x)^6 = 1 + 6(2x) + 15(2x)^2 + 20(2x)^3 + 15(2x)^4 + 6(2x)^5 + (2x)^6 \ $$ = 1 + 12x + 60x^2 + 160x^3 + 240x^4 + 192x^5 + 64x^6$.

$(1-x)^7 = 1 - 7x + 21x^2 - 35x^3 + 35x^4 - 21x^5 + \dots$.

To get the term with $x^5$ in the product, we multiply terms from each expansion whose powers of 'x' sum to 5:

• (Constant from 1st) $\times$ ($x^5$ from 2nd): $(1)(-21x^5) = -21x^5$.
• ($x^1$ from 1st) $\times$ ($x^4$ from 2nd): $(12x)(35x^4) = 420x^5$.
• ($x^2$ from 1st) $\times$ ($x^3$ from 2nd): $(60x^2)(-35x^3) = -2100x^5$.
• ($x^3$ from 1st) $\times$ ($x^2$ from 2nd): $(160x^3)(21x^2) = 3360x^5$.
• ($x^4$ from 1st) $\times$ ($x^1$ from 2nd): $(240x^4)(-7x) = -1680x^5$.
• ($x^5$ from 1st) $\times$ (Constant from 2nd): $(192x^5)(1) = 192x^5$.

The total coefficient of $x^5$ is the sum of these individual coefficients:

$-21 + 420 - 2100 + 3360 - 1680 + 192 = 171$.

The coefficient is 171.

To Prove:

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

Proof:

We write $a = (a - b) + b$.

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

Using the Binomial Theorem to expand this expression:

$a^n = {^nC_0}(a-b)^n + {^nC_1}(a-b)^{n-1}b + \dots + {^nC_{n-1}}(a-b)b^{n-1} + {^nC_n}b^n$.

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

Now, subtract $b^n$ from both sides:

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

Every term on the right-hand side has a factor of $(a-b)$. We can factor it out:

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

Let $K = \left[ (a-b)^{n-1} + n(a-b)^{n-2}b + \dots + nb^{n-1} \right]$. Since $a, b, n$ are integers, $K$ is also an integer.

So, $a^n - b^n = (a-b)K$.

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

Hence, proved.

To Evaluate:

$(\sqrt{3} + \sqrt{2})^6 - (\sqrt{3} - \sqrt{2})^6$.

Solution:

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

When we subtract the expansion of $(a-b)^6$ from $(a+b)^6$, the terms with even powers of $b$ cancel out, and the terms with odd powers of $b$ are doubled.

$(a+b)^6 - (a-b)^6 = 2 \left[ {^6C_1}a^5b^1 + {^6C_3}a^3b^3 + {^6C_5}a^1b^5 \right]$.

The coefficients are $^6C_1=6, ^6C_3=20, ^6C_5=6$.

$= 2 \left[ 6(\sqrt{3})^5(\sqrt{2}) + 20(\sqrt{3})^3(\sqrt{2})^3 + 6(\sqrt{3})(\sqrt{2})^5 \right]$.

$(\sqrt{3})^5 = 9\sqrt{3}$, $(\sqrt{3})^3=3\sqrt{3}$.

$(\sqrt{2})^5 = 4\sqrt{2}$, $(\sqrt{2})^3=2\sqrt{2}$.

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

$= 2 \left[ 54\sqrt{6} + 120\sqrt{6} + 24\sqrt{6} \right]$.

$= 2 \left[ (54+120+24)\sqrt{6} \right] = 2[198\sqrt{6}] = 396\sqrt{6}$.

Therefore, the value is $396\sqrt{6}$.

To Find:

The value of $(a^2 + \sqrt{a^2 - 1})^4 + (a^2 - \sqrt{a^2 - 1})^4$.

Solution:

Let $x = a^2$ and $y = \sqrt{a^2 - 1}$. The expression is in the form $(x+y)^4 + (x-y)^4$.

We use the Binomial Theorem to expand $(x+y)^4$ and $(x-y)^4$.

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

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

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

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

Now, substitute back $x=a^2$ and $y=\sqrt{a^2-1}$:

$= 2 \left[ (a^2)^4 + 6(a^2)^2(\sqrt{a^2-1})^2 + (\sqrt{a^2-1})^4 \right]$.

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

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

$= 2 \left[ a^8 + 6a^6 - 5a^4 - 2a^2 + 1 \right]$.

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

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

To Find:

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

Solution:

We write $0.99$ as $(1 - 0.01)$. The expression is $(1 - 0.01)^5$.

Using the Binomial Theorem for $(a-b)^n$ with $a=1, b=0.01, n=5$.

The first three terms are:

$T_1 = {^5C_0}(1)^5 = 1$.

$T_2 = -{^5C_1}(1)^4(0.01) = -5 \times 0.01 = -0.05$.

$T_3 = +{^5C_2}(1)^3(0.01)^2 = 10 \times 0.0001 = 0.001$.

The approximation is the sum of these three terms:

Approximation = $1 - 0.05 + 0.001 = 0.95 + 0.001 = 0.951$.

Therefore, the approximation is 0.951.

Given:

In the expansion of $\left( 2^{1/4} + 3^{-1/4} \right)^n$, the ratio of the 5th term from the beginning to the 5th term from the end is $\sqrt{6}:1$.

Solution:

Let $a = 2^{1/4}$ and $b = 3^{-1/4}$. The expansion is $(a+b)^n$.

5th term from beginning ($T_5$) = $^nC_4 a^{n-4} b^4$.

5th term from the end is the $(n-5+2) = (n-3)$-th term from the beginning. Its index is $r = n-4$.

Term = $^nC_{n-4} a^{n-(n-4)} b^{n-4} = {^nC_4} a^4 b^{n-4}$.

The ratio is $\frac{T_5(\text{begin})}{T_5(\text{end})} = \frac{^nC_4 a^{n-4} b^4}{^nC_4 a^4 b^{n-4}} = \frac{a^{n-8}}{b^{n-8}} = \left(\frac{a}{b}\right)^{n-8} = \sqrt{6}$.

$\frac{a}{b} = \frac{2^{1/4}}{3^{-1/4}} = 2^{1/4} \cdot 3^{1/4} = (2 \cdot 3)^{1/4} = 6^{1/4}$.

So, $\left(6^{1/4}\right)^{n-8} = 6^{1/2}$.

$6^{\frac{n-8}{4}} = 6^{\frac{1}{2}}$.

Equating the exponents: $\frac{n-8}{4} = \frac{1}{2} \implies n-8=2 \implies n=10$.

Therefore, $n=10$.

To Expand:

$\left(1+ \frac{x}{2} -\frac{2}{x}\right)^{4}$.

Solution:

We group the terms to apply the Binomial Theorem. Let $a = 1$ and $b = (\frac{x}{2} - \frac{2}{x})$.

The expansion is $(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$.

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

$= 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 + 6 - \frac{16}{x^2} + \frac{16}{x^4}\right)$.

$= 1 + 2x - \frac{8}{x} + \frac{3x^2}{2} - 12 + \frac{24}{x^2} + \frac{x^3}{2} - 6x + \frac{24}{x} - \frac{32}{x^3} + \frac{x^4}{16} - x^2 + 6 \ $$ - \frac{16}{x^2} + \frac{16}{x^4}$.

Combining like terms:

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

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

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

To Expand:

$(3x^2 – 2ax + 3a^2)^3$.

Solution:

Let $u = 3x^2 + 3a^2$ and $v = -2ax$. The expression is $(u+v)^3$.

$(u+v)^3 = u^3 + 3u^2v + 3uv^2 + v^3$.

Term 1: $u^3 = (3(x^2+a^2))^3 = 27(x^6 + 3x^4a^2 + 3x^2a^4 + a^6) = 27x^6 \ $$ + 81a^2x^4 + 81a^4x^2 + 27a^6$.

Term 2: $3u^2v = 3(3(x^2+a^2))^2(-2ax) = 3 \cdot 9(x^4+2x^2a^2+a^4)(-2ax) \ $$ = -54ax(x^4+2x^2a^2+a^4) = -54ax^5 - 108a^3x^3 - 54a^5x$.

Term 3: $3uv^2 = 3(3(x^2+a^2))(-2ax)^2 = 9(x^2+a^2)(4a^2x^2) \ $$ = 36a^2x^2(x^2+a^2) = 36a^2x^4 + 36a^4x^2$.

Term 4: $v^3 = (-2ax)^3 = -8a^3x^3$.

Combining all terms:

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

Grouping by powers of $x$:

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

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

The expansion is $27x^6 - 54ax^5 + 117a^2x^4 - 116a^3x^3 + 117a^4x^2 \ $$ - 54a^5x + 27a^6$.