Binomial theorem Exercise 14 Solution, Polytechnic 1st semester Maths Solution

 

Exercise 14 Solutions

1. Expand the following by the binomial theorem:

(a) $(x^2 + 2a)^5$

Ans.

Using the binomial theorem, $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$.

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

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

Calculating each term:

1. $\binom{5}{0}(x^2)^5(2a)^0 = 1 \cdot x^{10} \cdot 1 = x^{10}$
2. $\binom{5}{1}(x^2)^4(2a)^1 = 5 \cdot x^8 \cdot 2a = 10ax^8$
3. $\binom{5}{2}(x^2)^3(2a)^2 = 10 \cdot x^6 \cdot 4a^2 = 40a^2x^6$
4. $\binom{5}{3}(x^2)^2(2a)^3 = 10 \cdot x^4 \cdot 8a^3 = 80a^3x^4$
5. $\binom{5}{4}(x^2)^1(2a)^4 = 5 \cdot x^2 \cdot 16a^4 = 80a^4x^2$
6. $\binom{5}{5}(x^2)^0(2a)^5 = 1 \cdot 1 \cdot 32a^5 = 32a^5$

Final expansion:

$$(x^2 + 2a)^5 = x^{10} + 10ax^8 + 40a^2x^6 + 80a^3x^4 + 80a^4x^2 + 32a^5$$

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(b) $(x - a)^4$

Ans.

Here, $a = x$, $b = -a$, and $n = 4$.

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

Calculating each term:

1. $\binom{4}{0}x^4(-a)^0 = 1 \cdot x^4 \cdot 1 = x^4$
2. $\binom{4}{1}x^3(-a)^1 = 4 \cdot x^3 \cdot (-a) = -4ax^3$
3. $\binom{4}{2}x^2(-a)^2 = 6 \cdot x^2 \cdot a^2 = 6a^2x^2$
4. $\binom{4}{3}x(-a)^3 = 4 \cdot x \cdot (-a)^3 = -4a^3x$
5. $\binom{4}{4}(-a)^4 = 1 \cdot (-a)^4 = a^4$

Final expansion:

$$(x - a)^4 = x^4 - 4ax^3 + 6a^2x^2 - 4a^3x + a^4$$

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(c) $(x + 2y)^6$

Ans. 

Here, $a = x$, $b = 2y$, and $n = 6$.

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

Calculating each term:

1. $\binom{6}{0}x^6(2y)^0 = 1 \cdot x^6 \cdot 1 = x^6$
2. $\binom{6}{1}x^5(2y)^1 = 6 \cdot x^5 \cdot 2y = 12yx^5$
3. $\binom{6}{2}x^4(2y)^2 = 15 \cdot x^4 \cdot 4y^2 = 60y^2x^4$
4. $\binom{6}{3}x^3(2y)^3 = 20 \cdot x^3 \cdot 8y^3 = 160y^3x^3$
5. $\binom{6}{4}x^2(2y)^4 = 15 \cdot x^2 \cdot 16y^4 = 240y^4x^2$
6. $\binom{6}{5}x(2y)^5 = 6 \cdot x \cdot 32y^5 = 192y^5x$
7. $\binom{6}{6}(2y)^6 = 1 \cdot 64y^6 = 64y^6$

Final expansion:

$$(x + 2y)^6 = x^6 + 12yx^5 + 60y^2x^4 + 160y^3x^3 + 240y^4x^2 + 192y^5x + 64y^6$$

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

Ans. 

Here, $a = x$, $b = \frac{1}{x}$, and $n = 5$.

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

Calculating each term:

1. $\binom{5}{0}x^5\left(\frac{1}{x}\right)^0 = 1 \cdot x^5 \cdot 1 = x^5$
2. $\binom{5}{1}x^4\left(\frac{1}{x}\right)^1 = 5 \cdot x^4 \cdot \frac{1}{x} = 5x^3$
3. $\binom{5}{2}x^3\left(\frac{1}{x}\right)^2 = 10 \cdot x^3 \cdot \frac{1}{x^2} = 10x$
4. $\binom{5}{3}x^2\left(\frac{1}{x}\right)^3 = 10 \cdot x^2 \cdot \frac{1}{x^3} = \frac{10}{x}$
5. $\binom{5}{4}x\left(\frac{1}{x}\right)^4 = 5 \cdot x \cdot \frac{1}{x^4} = \frac{5}{x^3}$
6. $\binom{5}{5}\left(\frac{1}{x}\right)^5 = 1 \cdot \frac{1}{x^5} = \frac{1}{x^5}$

Final expansion:

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

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(e) $(3x + 2y)^4$

Ans. 

Using the binomial theorem, $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$.

Here, $a = 3x$, $b = 2y$, and $n = 4$.

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

Calculating each term:

1. $\binom{4}{0}(3x)^4(2y)^0 = 1 \cdot 81x^4 \cdot 1 = 81x^4$
2. $\binom{4}{1}(3x)^3(2y)^1 = 4 \cdot 27x^3 \cdot 2y = 216yx^3$
3. $\binom{4}{2}(3x)^2(2y)^2 = 6 \cdot 9x^2 \cdot 4y^2 = 216y^2x^2$
4. $\binom{4}{3}(3x)^1(2y)^3 = 4 \cdot 3x \cdot 8y^3 = 96y^3x$
5. $\binom{4}{4}(2y)^4 = 1 \cdot 16y^4 = 16y^4$

Final expansion:

$$(3x + 2y)^4 = 81x^4 + 216yx^3 + 216y^2x^2 + 96y^3x + 16y^4$$

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(f) $\left(\frac{\sqrt{x}}{\sqrt{a}} - \frac{\sqrt{a}}{\sqrt{x}}\right)^6$

Ans. 

Here, $a = \frac{\sqrt{x}}{\sqrt{a}}$, $b = -\frac{\sqrt{a}}{\sqrt{x}}$, and $n = 6$.

$$\left(\frac{\sqrt{x}}{\sqrt{a}} - \frac{\sqrt{a}}{\sqrt{x}}\right)^6 = \binom{6}{0}\left(\frac{\sqrt{x}}{\sqrt{a}}\right)^6 \left(-\frac{\sqrt{a}}{\sqrt{x}}\right)^0 + \binom{6}{1}\left(\frac{\sqrt{x}}{\sqrt{a}}\right)^5 \left(-\frac{\sqrt{a}}{\sqrt{x}}\right)^1 + \ldots + \binom{6}{6}\left(\frac{\sqrt{a}}{\sqrt{x}}\right)^0 \left(-\frac{\sqrt{a}}{\sqrt{x}}\right)^6$$

Calculating each term:

1. $\binom{6}{0}\left(\frac{\sqrt{x}}{\sqrt{a}}\right)^6 = \left(\frac{x}{a}\right)^3 = \frac{x^3}{a^3}$
2. $\binom{6}{1}\left(\frac{\sqrt{x}}{\sqrt{a}}\right)^5 \cdot \left(-\frac{\sqrt{a}}{\sqrt{x}}\right) = -6\frac{x^{5/2} \cdot a^{1/2}}{a^{5/2} \cdot x^{1/2}} = -6\frac{x^2}{a^2}$
3. $\binom{6}{2}\left(\frac{\sqrt{x}}{\sqrt{a}}\right)^4 \cdot \left(-\frac{\sqrt{a}}{\sqrt{x}}\right)^2 = 15\frac{x^2}{a^2} = 15\frac{x}{a}$
4. $\binom{6}{3}\left(\frac{\sqrt{x}}{\sqrt{a}}\right)^3 \cdot \left(-\frac{\sqrt{a}}{\sqrt{x}}\right)^3 = -20$
5. $\binom{6}{4}\left(\frac{\sqrt{x}}{\sqrt{a}}\right)^2 \cdot \left(-\frac{\sqrt{a}}{\sqrt{x}}\right)^4 = 15\frac{a}{x}$
6. $\binom{6}{5}\left(\frac{\sqrt{x}}{\sqrt{a}}\right)^1 \cdot \left(-\frac{\sqrt{a}}{\sqrt{x}}\right)^5 = -6\frac{a^2}{x^2}$
7. $\binom{6}{6}\left(\frac{\sqrt{a}}{\sqrt{x}}\right)^6 = \frac{a^3}{x^3}$

Final expansion:

$$\left(\frac{\sqrt{x}}{\sqrt{a}} - \frac{\sqrt{a}}{\sqrt{x}}\right)^6 = \frac{x^3}{a^3} - 6\frac{x^2}{a^2} + 15\frac{x}{a} - 20 + 15\frac{a}{x} - 6\frac{a^2}{x^2} + \frac{a^3}{x^3}$$

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2. Find the value of the following with the help of the binomial theorem:

(a) $(1 + \sqrt{5})^5 + (1 - \sqrt{5})^5$

Let’s denote $(1 + \sqrt{5}) = a$ and $(1 - \sqrt{5}) = b$.

We need to find $a^5 + b^5$.

Using the identity:

$$a^5 + b^5 = (a + b)(a^4 - a^3b + a^2b^2 - ab^3 + b^4)$$

Here, $a + b = 2$ and $ab = (1 + \sqrt{5})(1 - \sqrt{5}) = 1 - 5 = -4$.

Substituting $a + b = 2$ and $ab = -4$, the expression simplifies to:

$$(2)\left(2^4 + 10(4)\right) = 0$$

Hence, $(1 + \sqrt{5})^5 + (1 - \sqrt{5})^5 = 0$.

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

Similarly, let $a = \sqrt{2} + \sqrt{3}$ and $b = \sqrt{2} - \sqrt{3}$.

We need to find $a^6 + b^6$.

Using the identity:

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

Calculating $a^2 + b^2$:

$$a^2 + b^2 = (\sqrt{2} + \sqrt{3})^2 + (\sqrt{2} - \sqrt{3})^2 = 2 + 2\sqrt{6} + 3 + 2 - 2\sqrt{6} + 3 = 10$$

Calculating $a^2b^2$:

$$a^2b^2 = (2 - \sqrt{6}) \cdot (2 + \sqrt{6}) = 10$$

Final result:

$$(\sqrt{2} + \sqrt{3})^6 + (\sqrt{2} - \sqrt{3})^6 = 10$$

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(c) $(1 + 2\sqrt{5})^5 + (1 - 2\sqrt{5})^5$

Let’s denote $a = (1 + 2\sqrt{5})$ and $b = (1 - 2\sqrt{5})$.

We need to find $a^5 + b^5$.

Using the identity:

$$a^5 + b^5 = (a + b)(a^4 - a^3b + a^2b^2 - ab^3 + b^4)$$

Calculating $a + b$ and $ab$:

$$a + b = (1 + 2\sqrt{5}) + (1 - 2\sqrt{5}) = 2$$ $$ab = (1 + 2\sqrt{5})(1 - 2\sqrt{5}) = 1 - (2\sqrt{5})^2 = 1 - 20 = -19$$

Using this, we can rewrite the identity as:

$$a^5 + b^5 = 2\left(a^4 - a^3b + a^2b^2 - ab^3 + b^4\right)$$

We notice that all odd power terms in $a$ and $b$ will cancel out due to symmetry, so the final answer will involve only even terms. Simplifying, we get:

$$(1 + 2\sqrt{5})^5 + (1 - 2\sqrt{5})^5 = 2\left(1 + 5 + 50 + 2500 \right) = 2 \times 2556 = 5112$$

Hence, $(1 + 2\sqrt{5})^5 + (1 - 2\sqrt{5})^5 = 5112$.

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(d) $(3 + \sqrt{2})^5 + (3 - \sqrt{2})^5$

Let’s denote $a = (3 + \sqrt{2})$ and $b = (3 - \sqrt{2})$.

We need to find $a^5 + b^5$.

Using the identity:

$$a^5 + b^5 = (a + b)(a^4 - a^3b + a^2b^2 - ab^3 + b^4)$$

Calculating $a + b$ and $ab$:

$$a + b = (3 + \sqrt{2}) + (3 - \sqrt{2}) = 6$$ $$ab = (3 + \sqrt{2})(3 - \sqrt{2}) = 9 - 2 = 7$$

So the simplified expression becomes:

$$(3 + \sqrt{2})^5 + (3 - \sqrt{2})^5 = 6\left(1 + 5 \cdot 7 + 10 \cdot 7^2 + 10 \cdot 7^3 + 5 \cdot 7^4\right)$$

Calculating step-by-step:

$$6(1 + 35 + 490 + 3430 + 12005) = 6 \times 15961 = 95766$$

Hence, $(3 + \sqrt{2})^5 + (3 - \sqrt{2})^5 = 95766$.

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3. Find the value of the following with the help of the binomial theorem:

(a) $(101)^4$

Ans. 

We can express $101$ as $(100 + 1)$.

Using the binomial theorem:

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

Calculating each term:

1. $\binom{4}{0}100^4 \cdot 1 = 100000000$
2. $\binom{4}{1}100^3 \cdot 1 = 4000000$
3. $\binom{4}{2}100^2 \cdot 1 = 60000$
4. $\binom{4}{3}100^1 \cdot 1 = 400$
5. $\binom{4}{4}1 = 1$

Adding all the terms:

$$(100 + 1)^4 = 100000000 + 4000000 + 60000 + 400 + 1 = 104060401$$

Hence, $(101)^4 = 104060401$.

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(b) $(99)^5$

Ans. 

We can express $99$ as $(100 - 1)$.

Using the binomial theorem:

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

Calculating each term:

1. $\binom{5}{0}100^5 \cdot 1 = 10000000000$
2. $\binom{5}{1}100^4 \cdot (-1) = -500000000$
3. $\binom{5}{2}100^3 \cdot 1 = 10000000$
4. $\binom{5}{3}100^2 \cdot (-1) = -100000$
5. $\binom{5}{4}100^1 \cdot 1 = 500$
6. $\binom{5}{5}(-1) = -1$

Adding all the terms:

$$(100 - 1)^5 = 10000000000 - 500000000 + 10000000 - 100000 + 500 - 1 = 9500999499$$

Hence, $(99)^5 = 9500999499$.

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(c) $(10.1)^5$

Ans. 

We can express $10.1$ as $(10 + 0.1)$.

Using the binomial theorem:

$$(10 + 0.1)^5 = \binom{5}{0}10^5 \cdot (0.1)^0 + \binom{5}{1}10^4 \cdot (0.1)^1 + \binom{5}{2}10^3 \cdot (0.1)^2 + \binom{5}{3}10^2 \cdot (0.1)^3 + \binom{5}{4}10^1 \cdot (0.1)^4 + \binom{5}{5}(0.1)^5$$

Calculating each term:

1. $\binom{5}{0}10^5 \cdot (0.1)^0 = 100000$
2. $\binom{5}{1}10^4 \cdot (0.1)^1 = 5000$
3. $\binom{5}{2}10^3 \cdot (0.1)^2 = 100$
4. $\binom{5}{3}10^2 \cdot (0.1)^3 = 1$
5. $\binom{5}{4}10^1 \cdot (0.1)^4 = 0.005$
6. $\binom{5}{5}(0.1)^5 = 0.00001$

Adding all the terms:

$$(10.1)^5 = 100000 + 5000 + 100 + 1 + 0.005 + 0.00001 = 105101.00501$$

Hence, $(10.1)^5 = 105101.00501$.

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(d) $(1.1)^4$

Ans.

We can express $1.1$ as $(1 + 0.1)$.

Using the binomial theorem:

$$(1 + 0.1)^4 = \binom{4}{0}(1)^4 \cdot (0.1)^0 + \binom{4}{1}(1)^3 \cdot (0.1)^1 + \binom{4}{2}(1)^2 \cdot (0.1)^2 + \binom{4}{3}(1)^1 \cdot (0.1)^3 + \binom{4}{4}(0.1)^4$$

Calculating each term:

1. $\binom{4}{0}(1)^4 \cdot (0.1)^0 = 1$
2. $\binom{4}{1}(1)^3 \cdot (0.1)^1 = 0.4$
3. $\binom{4}{2}(1)^2 \cdot (0.1)^2 = 0.06$
4. $\binom{4}{3}(1)^1 \cdot (0.1)^3 = 0.004$
5. $\binom{4}{4}(0.1)^4 = 0.0001$

Adding all the terms:

$$(1 + 0.1)^4 = 1 + 0.4 + 0.06 + 0.004 + 0.0001 = 1.4641$$

Hence, $(1.1)^4 = 1.4641$.

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4. Find the indicated term in the expansion of the following:

(a) 5th term of $(a + 2x^3)^{17}$

Ans.

The general term in the expansion of $(a + 2x^3)^{17}$ is given by:

$$T_{r+1} = \binom{17}{r} a^{17-r} (2x^3)^r$$

We need to find the 5th term, which means $r = 4$.

Substitute $r = 4$ into the general formula:

$$T_5 = \binom{17}{4} a^{17-4} (2x^3)^4$$

Calculating step-by-step:

1. $\binom{17}{4} = \frac{17 \times 16 \times 15 \times 14}{4 \times 3 \times 2 \times 1} = 2380$
2. $a^{17-4} = a^{13}$
3. $(2x^3)^4 = 2^4 \times (x^3)^4 = 16x^{12}$

Combining these, we get:

$$T_5 = 2380 \times a^{13} \times 16x^{12} = 38080 \, a^{13} \, x^{12}$$

Hence, the 5th term is $38080 \, a^{13} \, x^{12}$.

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(b) 5th term of $(x - 2)^8$

Ans.

The general term in the expansion of $(x - 2)^8$ is given by:

$$T_{r+1} = \binom{8}{r} x^{8-r} (-2)^r$$

We need to find the 5th term, which means $r = 4$.

Substitute $r = 4$ into the general formula:

$$T_5 = \binom{8}{4} x^{8-4} (-2)^4$$

Calculating step-by-step:

1. $\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$
2. $x^{8-4} = x^{4}$
3. $(-2)^4 = 16$

Combining these, we get:

$$T_5 = 70 \times x^{4} \times 16 = 1120 \, x^{4}$$

Hence, the 5th term is $1120 \, x^{4}$.

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(c) 6th term of $\left(\frac{2}{x} - \frac{x^2}{2}\right)^9$

Solution.

The general term in the expansion of $\left(\frac{2}{x} - \frac{x^2}{2}\right)^9$ is given by:

$$T_{r+1} = \binom{9}{r} \left(\frac{2}{x}\right)^{9-r} \left(-\frac{x^2}{2}\right)^r$$

We need to find the 6th term, which means $r = 5$.

Substitute $r = 5$ into the general formula:

$$T_6 = \binom{9}{5} \left(\frac{2}{x}\right)^{9-5} \left(-\frac{x^2}{2}\right)^5$$

Calculating step-by-step:

1. $\binom{9}{5} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$
2. $\left(\frac{2}{x}\right)^{9-5} = \left(\frac{2}{x}\right)^{4} = \frac{16}{x^{4}}$
3. $\left(-\frac{x^2}{2}\right)^5 = \frac{(-1)^5 \times x^{10}}{32} = -\frac{x^{10}}{32}$

Combining these, we get:

$$T_6 = 126 \times \frac{16}{x^{4}} \times \left(-\frac{x^{10}}{32}\right)$$

Simplifying further:

$$T_6 = 126 \times \left(-\frac{1}{2}\right) \times x^{6} = -63 \, x^{6}$$

Hence, the 6th term is $-63 \, x^{6}$.

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(d) 10th term of $\left(x - \frac{1}{x}\right)^{12}$

Solution.

The general term in the expansion of $\left(x - \frac{1}{x}\right)^{12}$ is given by:

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

We need to find the 10th term, which means $r = 9$.

Substitute $r = 9$ into the general formula:

$$T_{10} = \binom{12}{9} x^{12-9} \left(-\frac{1}{x}\right)^9$$

Calculating step-by-step:

1. $\binom{12}{9} = \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$
2. $x^{12-9} = x^{3}$
3. $\left(-\frac{1}{x}\right)^9 = \frac{(-1)^9}{x^9} = -\frac{1}{x^9}$

Combining these, we get:

$$T_{10} = 220 \times x^{3} \times \left(-\frac{1}{x^9}\right) = -220 \times \frac{x^{3}}{x^{9}} = -220 \times \frac{1}{x^{6}} = -\frac{220}{x^{6}}$$

Hence, the 10th term is $-\frac{220}{x^{6}}$.

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(e) 25th term of $\left(\sqrt{\frac{x}{y}} - \frac{y}{\sqrt{x}}\right)^{26}$

Solution.

The general term in the expansion of $\left(\sqrt{\frac{x}{y}} - \frac{y}{\sqrt{x}}\right)^{26}$ is given by:

$$T_{r+1} = \binom{26}{r} \left(\sqrt{\frac{x}{y}}\right)^{26-r} \left(-\frac{y}{\sqrt{x}}\right)^r$$

We need to find the 25th term, which means $r = 24$.

Substitute $r = 24$ into the general formula:

$$T_{25} = \binom{26}{24} \left(\sqrt{\frac{x}{y}}\right)^{26-24} \left(-\frac{y}{\sqrt{x}}\right)^{24}$$

Calculating step-by-step:

1. $\binom{26}{24} = \binom{26}{2} = \frac{26 \times 25}{2} = 325$
2. $\left(\sqrt{\frac{x}{y}}\right)^{26-24} = \left(\sqrt{\frac{x}{y}}\right)^{2} = \frac{x}{y}$
3. $\left(-\frac{y}{\sqrt{x}}\right)^{24} = \frac{y^{24}}{x^{12}}$ (since $(-1)^{24} = 1$)

Combining these, we get:

$$T_{25} = 325 \times \frac{x}{y} \times \frac{y^{24}}{x^{12}} = 325 \times \frac{y^{23}}{x^{11}}$$

Hence, the 25th term is $\frac{325 \, y^{23}}{x^{11}}$.

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5. Find the middle terms in the expansion of the following:

To find the middle term(s) in the expansion of a binomial expression, we first need to determine the number of terms in the expansion and then identify the middle one or two terms depending on whether the total number of terms is odd or even.

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(a) $\left(x - \frac{1}{x}\right)^{10}$

Solution.

The number of terms in this expansion is $10 + 1 = 11$.

Since the number of terms is odd, there will be a single middle term, which is the $6^{th}$ term.

The general term in the expansion is:

$$T_{r+1} = \binom{10}{r} x^{10-r} \left(-\frac{1}{x}\right)^r$$

For the middle term, $r = 5$:

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

Calculating step-by-step:

1. $\binom{10}{5} = 252$
2. $x^{10-5} = x^{5}$
3. $\left(-\frac{1}{x}\right)^5 = \frac{(-1)^5}{x^5} = -\frac{1}{x^5}$

Combining these, we get:

$$T_6 = 252 \times x^{5} \times \left(-\frac{1}{x^5}\right) = -252$$

Hence, the middle term is $-252$.

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(b) $\left(\frac{a}{3} - \frac{b}{2}\right)^{6}$

Solution.

The number of terms in this expansion is $6 + 1 = 7$.

Since the number of terms is odd, there will be a single middle term, which is the $4^{th}$ term.

The general term in the expansion is:

$$T_{r+1} = \binom{6}{r} \left(\frac{a}{3}\right)^{6-r} \left(-\frac{b}{2}\right)^r$$

For the middle term, $r = 3$:

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

Calculating step-by-step:

1. $\binom{6}{3} = 20$
2. $\left(\frac{a}{3}\right)^{3} = \frac{a^{3}}{27}$
3. $\left(-\frac{b}{2}\right)^3 = \frac{-b^{3}}{8}$

Combining these, we get:

$$T_4 = 20 \times \frac{a^{3}}{27} \times \frac{-b^{3}}{8} = \frac{-20 a^{3} b^{3}}{216}$$

Simplifying further:

$$T_4 = -\frac{5 a^{3} b^{3}}{54}$$

Hence, the middle term is $-\frac{5 a^{3} b^{3}}{54}$.

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(c) $(y + x)^{11}$

Solution.

The number of terms in this expansion is $11 + 1 = 12$.

Since the number of terms is even, there will be two middle terms, the $6^{th}$ and $7^{th}$ terms.

6th Term:

The general term is:

$$T_{r+1} = \binom{11}{r} y^{11-r} x^r$$

For the 6th term, $r = 5$:

$$T_6 = \binom{11}{5} y^{11-5} x^{5}$$

Calculating step-by-step:

1. $\binom{11}{5} = 462$
2. $y^{11-5} = y^{6}$
3. $x^{5} = x^{5}$

Combining these, we get:

$$T_6 = 462 \, y^{6} \, x^{5}$$

Hence, the 6th term is $462 \, y^{6} \, x^{5}$.

7th Term:

For the 7th term, $r = 6$:

$$T_7 = \binom{11}{6} y^{11-6} x^{6}$$

Calculating step-by-step:

1. $\binom{11}{6} = 462$
2. $y^{11-6} = y^{5}$
3. $x^{6} = x^{6}$

Combining these, we get:

$$T_7 = 462 \, y^{5} \, x^{6}$$

Hence, the 7th term is $462 \, y^{5} \, x^{6}$.

The middle terms are:

$$462 \, y^{6} \, x^{5} \quad \text{and} \quad 462 \, y^{5} \, x^{6}$$

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(d) $\left(3a - \frac{a^3}{6}\right)^{9}$

Ans. 

The number of terms in this expansion is $9 + 1 = 10$.

Since the number of terms is even, there will be two middle terms, the $5^{th}$ and $6^{th}$ terms.

5th Term:

The general term is:

$$T_{r+1} = \binom{9}{r} \left(3a\right)^{9-r} \left(-\frac{a^3}{6}\right)^r$$

For the 5th term, $r = 4$:

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

Calculating step-by-step:

1. $\binom{9}{4} = 126$
2. $\left(3a\right)^{5} = 243a^5$
3. $\left(-\frac{a^3}{6}\right)^4 = \frac{a^{12}}{1296}$

Combining these, we get:

$$T_5 = 126 \times 243a^5 \times \frac{a^{12}}{1296} = \frac{30618a^{17}}{1296}$$

Simplifying further:

$$T_5 = \frac{1701a^{17}}{72}$$

6th Term:

For the 6th term, $r = 5$:

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

Calculating step-by-step:

1. $\binom{9}{5} = 126$
2. $\left(3a\right)^{4} = 81a^4$
3. $\left(-\frac{a^3}{6}\right)^5 = \frac{-a^{15}}{7776}$

Combining these, we get:

$$T_6 = 126 \times 81a^4 \times \frac{-a^{15}}{7776} = \frac{-10206a^{19}}{7776}$$

Simplifying further:

$$T_6 = \frac{-1701a^{19}}{1296}$$

The middle terms are:

$$\frac{1701a^{17}}{72} \quad \text{and} \quad \frac{-1701a^{19}}{1296}$$

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6. Expand the following up to 4 terms:

In this question, we need to expand the given expressions using the Binomial Theorem up to the first four terms.

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

Solution.

The general binomial expansion for $(1 + x)^n$ up to four terms is:

$$1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots$$

In this case, $n = \frac{1}{2}$ and $x = -\frac{x}{2}$.

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

Calculating step-by-step:

1. First term: $1$

2. Second term: $\frac{1}{2} \times \left(-\frac{x}{2}\right) = -\frac{x}{4}$

3. Third term:

   $$\frac{\frac{1}{2} \left(\frac{1}{2} - 1\right)}{2} \times \left(-\frac{x}{2}\right)^2 = \frac{\frac{1}{2} \times \left(-\frac{1}{2}\right)}{2} \times \frac{x^2}{4} = \frac{-1/4}{8} x^2 = -\frac{x^2}{32}$$

4. Fourth term:

   $$\frac{\frac{1}{2} \left(\frac{1}{2} - 1\right) \left(\frac{1}{2} - 2\right)}{6} \times \left(-\frac{x}{2}\right)^3 = \frac{\frac{1}{2} \times \left(-\frac{1}{2}\right) \times \left(-\frac{3}{2}\right)}{6} \times \frac{-x^3}{8}$$ $$= \frac{-3/16}{48} x^3 = \frac{3 x^3}{384} = \frac{x^3}{128}$$

Combining these, we get the expansion up to four terms:

$$\left(1 - \frac{x}{2}\right)^{\frac{1}{2}} = 1 - \frac{x}{4} - \frac{x^2}{32} + \frac{x^3}{128}$$

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(b) $(8 + x)^{\frac{4}{3}}$

Solutoin.

We will use the binomial theorem for $(a + x)^n$. Let’s rewrite it as $(8(1 + \frac{x}{8}))^{\frac{4}{3}}$.

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

Using the binomial expansion for $(1 + x)^n$ up to four terms:

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

Calculating step-by-step:

1. First term: $1$

2. Second term: $\frac{4}{3} \times \frac{x}{8} = \frac{x}{6}$

3. Third term:

   $$\frac{\frac{4}{3} \times \frac{1}{3}}{2} \times \frac{x^2}{64} = \frac{2/9}{2} \times \frac{x^2}{64} = \frac{x^2}{576}$$

4. Fourth term:

   $$\frac{\frac{4}{3} \times \frac{1}{3} \times \left(-\frac{2}{3}\right)}{6} \times \frac{x^3}{512}$$ $$= \frac{-8/81}{6} \times \frac{x^3}{512} = \frac{-x^3}{41472}$$

Combining these, we get:

$$\left(1 + \frac{x}{8}\right)^{\frac{4}{3}} = 1 + \frac{x}{6} + \frac{x^2}{576} - \frac{x^3}{41472}$$

Now, multiplying by $16$:

$$(8 + x)^{\frac{4}{3}} = 16 \left(1 + \frac{x}{6} + \frac{x^2}{576} - \frac{x^3}{41472}\right)$$ $$= 16 + \frac{16x}{6} + \frac{16x^2}{576} - \frac{16x^3}{41472}$$

Simplifying further:

$$(8 + x)^{\frac{4}{3}} = 16 + \frac{8x}{3} + \frac{x^2}{36} - \frac{x^3}{2592}$$

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(c) $(3 - 2x)^{-\frac{2}{3}}$

Solution.

We will use the binomial expansion for $(a + x)^n$ where $n$ is a negative fraction. Let’s rewrite it as:

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

Using the binomial expansion for $(1 + x)^n$ up to four terms:

$$1 - n \frac{2x}{3} + \frac{n(n-1)}{2} \left(\frac{2x}{3}\right)^2 - \frac{n(n-1)(n-2)}{6} \left(\frac{2x}{3}\right)^3 + \dots$$

In this case, $n = -\frac{2}{3}$.

Calculating step-by-step:

1. First term: $1$

2. Second term:

   $$-\left(-\frac{2}{3}\right) \times \frac{2x}{3} = \frac{4x}{9}$$

3. Third term:

   $$\frac{-\frac{2}{3} \times \left(-\frac{5}{3}\right)}{2} \times \frac{4x^2}{9} = \frac{5 \times 4 x^2}{81} = \frac{20 x^2}{81}$$

4. Fourth term:

   $$\frac{-\frac{2}{3} \times \left(-\frac{5}{3}\right) \times \left(-\frac{8}{3}\right)}{6} \times \frac{8 x^3}{27}$$ $$= \frac{-80}{729} x^3$$

Combining these, we get:

$$\left(1 - \frac{2x}{3}\right)^{-\frac{2}{3}} = 1 + \frac{4x}{9} + \frac{20 x^2}{81} - \frac{80 x^3}{729}$$

Now, multiplying by $3^{-\frac{2}{3}}$:

$$(3 - 2x)^{-\frac{2}{3}} = 3^{-\frac{2}{3}} \left(1 + \frac{4x}{9} + \frac{20 x^2}{81} - \frac{80 x^3}{729}\right)$$

$$Que\; 7. \; find\; the\; term\; independent\; of\; x\; in\; the\; expansion\; of\; (3x2/2-1/3x)6$$

To find the term independent of $x$ in the expansion of $\left( \frac{3x^2}{2} - \frac{1}{3x} \right)^6$, we need to find the term where the power of $x$ becomes zero.

Step-by-Step Solution:

1. Identify the General Term: The general term in the expansion of $\left( \frac{3x^2}{2} - \frac{1}{3x} \right)^6$ is given by:

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

   Simplifying, we get:

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

2. Condition for Term Independent of $x$:

   The term will be independent of $x$ when the power of $x$ is zero, i.e., $12 - 3r = 0$.

   Solving for $r$:

   $$12 - 3r = 0$$ $$3r = 12$$ $$r = 4$$

3. Find the Term $T_5$:

   Now, substitute $r = 4$ into the general term formula:

   $$T_{4+1} = \binom{6}{4} \frac{(-1)^4 \cdot 3^{6-2 \times 4} \cdot x^{12-3 \times 4}}{2^{6-4}}$$

   Simplifying:

   $$T_5 = \binom{6}{4} \frac{1 \cdot 3^{6-8} \cdot x^{0}}{2^2}$$ $$T_5 = \binom{6}{4} \frac{3^{-2}}{4}$$

   Since $\binom{6}{4} = 15$ and $3^{-2} = \frac{1}{9}$, we have:

   $$T_5 = \frac{15 \times 1}{9 \times 4}$$ $$T_5 = \frac{15}{36}$$

   Simplify further:

   $$T_5 = \frac{5}{12}$$

Therefore, the term independent of $x$ in the expansion of $\left( \frac{3x^2}{2} - \frac{1}{3x} \right)^6$ is $\frac{5}{12}$.