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NCERT Exemplar for Class 11 Maths Chapter 8 - Binomial Theorem (Book Solutions)

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NCERT Exemplar for Class 11 Maths Chapter 8- Binomial Theorem - Free PDF Download

Download free PDF of NCERT Exemplar for Class 11 Maths Chapter 8 - Binomial Theorem solved by expert Maths teachers of Vedantu.com by following the NCERT (CBSE) Book guidelines. All Chapter 8 - Binomial Theorem exercises questions with solutions will help you to revise and complete the syllabus and score more marks in your examinations.

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Access NCERT Exemplar Solutions for Class 11 Mathematics Chapter 8: Binomial Theorem (Examples, Easy Methods and Step by Step Solutions)

Sample Question 1: Find the $\mathrm{r}^{\mathrm{th}}$ term in expansion of $\left(x+\dfrac{1}{x}\right)^{2 r}$

Ans: Given, $\left(\mathrm{x}+\dfrac{1}{\mathrm{x}}\right)^{2 \mathrm{r}}$

Use the general formula,

$\mathrm{T}_{\mathrm{r+1}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(\mathrm{a})^{\mathrm{n}-\mathrm{r}} \mathrm{~b}^{\mathrm{r}}$ in expansion $(\mathrm{a}+\mathrm{b})^{\mathrm{n}}$

$\mathrm{r} \rightarrow \mathrm{r}-1$

$\Rightarrow \mathrm{T}_{\mathrm{r}}={ }^{2 \mathrm{r}} \mathrm{C}_{\mathrm{r}-1}(\mathrm{x})^{2 \mathrm{r}-\mathrm{r}+1}\left(\dfrac{1}{\mathrm{x}}\right)^{\mathrm{r}-1}$

$=\dfrac{2 \mathrm{r} !}{(\mathrm{r}-1) !(\mathrm{r}+1) !} \mathrm{x}^{\mathrm{r+1-\mathrm{r}+1}}$

$=\dfrac{2 \mathrm{r} !}{(\mathrm{r}-1) !(\mathrm{r}+1) !} \mathrm{x}^{2}$

Sample Question 2: Expand the following $\left(1-x+x^{2}\right)^{4}$.

Ans: Given, $\left(1-\mathrm{x}+\mathrm{x}^{2}\right)^{4}$.

Use $\mathrm{y}=1-\mathrm{x}$. Then, use general binomial expansion.

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

$={ }^{4} \mathrm{C}_{0} \mathrm{y}^{4}\left(\mathrm{x}^{2}\right)^{0}+{ }^{4} \mathrm{C}_{1} \mathrm{y}^{3}\left(\mathrm{x}^{2}\right)^{1}+{ }^{4} \mathrm{C}_{2} \mathrm{y}^{2}\left(\mathrm{x}^{2}\right)^{2}+{ }^{4} \mathrm{C}_{3} \mathrm{y}^{1}\left(\mathrm{x}^{2}\right)^{3}+{ }^{4} \mathrm{C}_{4} \mathrm{y}^{0}\left(\mathrm{x}^{2}\right)^{4}$

$=\mathrm{y}^{4}+4 \mathrm{y}^{3} \mathrm{x}^{2}+6 \mathrm{y}^{2} \mathrm{x}^{4}+4 \mathrm{yx}^{6}+\mathrm{x}^{8}$

$=(1-\mathrm{x})^{4}+4 \mathrm{x}^{2}(1-\mathrm{x})^{3}+6 x^{4}(1-\mathrm{x})^{2}+4 x^{6}(1-\mathrm{x})+\mathrm{x}^{8}$

$=1-4 x+10 x^{2}-16 x^{8}+10 x^{4}-16 x^{8}+10 x^{6}-4 x^{7}+x^{4}$

3. Find the $4^{\text {th }}$ term from end in expansion of $\left(\dfrac{x^{3}}{2}-\dfrac{2}{x^{2}}\right)^{9}$.

Ans: Given, $\left(\dfrac{x^{3}}{2}-\dfrac{2}{x^{2}}\right)^{9}$.

$r^{\text {th }}$ term from end in expansion of $(a+b)^{n}$ is $(n-r+2)^{\text {th }}$ term from beginning.

Here, $4^{\text {th }}$ term from end is $(9-4+2)^{\text {th }}=7^{\text {th }}$ term from beginning.

$\mathrm{T}_{7}={ }^{9} \mathrm{C}_{6}\left(\dfrac{\mathrm{x}^{3}}{2}\right)^{3}\left(-\dfrac{2}{\mathrm{x}^{2}}\right)^{6}$

$={ }^{9} \mathrm{C}_{3} \cdot \dfrac{\mathrm{x}^{9}}{8} \cdot \dfrac{64}{\mathrm{x}^{12}}$

$=\dfrac{9 \times 8 \times 7}{3 \times 2 \times 1} \times \dfrac{8}{\mathrm{x}^{3}}$

$=\dfrac{672}{\mathrm{x}^{3}}$

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

Ans: Given, $\left(x^{2}-\sqrt{1-x^{2}}\right)^{4}+\left(x^{2}+\sqrt{1-x^{2}}\right)^{4}$.

Substitute $\sqrt{1-x^{2}}=y$. Then, solve.

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

$=2\left[x^{8}+{ }^{4} C_{2} x^{4} y^{2}+{ }^{4} C_{4} x^{4} y^{4}\right]$

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

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

5. Find the coefficient of $x^{11}$ in expansion of $\left(x^{3}-\dfrac{2}{x^{2}}\right)^{12}$.

Ans: Given, $\left(x^{3}-\dfrac{2}{x^{2}}\right)^{12}$

Consider the general $x^{11}$ is contained in general term.

$T_{r+1}={ }^{12} C_{r}\left(x^{3}\right)^{12-r}\left(-\dfrac{2}{x^{2}}\right)^{r}$

$={ }^{12} C_{r} x^{36-3 r-2 r}(-1)^{r} . 2^{r}$

$={ }^{12} C_{r}(-1)^{r} 2^{r} x^{36-5 r}$

$\text { for, } x^{11}=x^{36-5 r}$

$36-5 r=11$

$\Rightarrow r=5$

$\therefore$ Coefficient of $\mathrm{x}^{11}$ is

${ }^{12} \mathrm{C}_{5}(-1)^{5} 2^{5}=-\dfrac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2} \times 32$

$=-25344$

6. Determine whether the expansion of $\left(x^{2}-\dfrac{2}{x}\right)^{18}$ will contain a term containing $x^{10}$ ?

Ans: Given, $\left(x^{2}-\dfrac{2}{x}\right)^{18}$

In general term, $r$ cannot be a fraction.

$\mathrm{T}_{\mathrm{r}+1}={ }^{18} \mathrm{C}_{\mathrm{r}}\left(\mathrm{x}^{2}\right)^{18-\mathrm{r}}\left(-\dfrac{2}{\mathrm{x}}\right)^{\mathrm{r}}$

$={ }^{18} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{36-2 \mathrm{r}}(-1)^{\mathrm{r}} 2^{\mathrm{r}} \mathrm{x}^{-\mathrm{r}}$

$={ }^{18} \mathrm{C}_{\mathrm{r}}(-1)^{\mathrm{r}} 2^{\mathrm{r}} \mathrm{x}^{36-3 \mathrm{r}}$

$\text { Here, } 36-3 \mathrm{r}=10$

$\Rightarrow \mathrm{r}=\dfrac{26}{3} \text { (not possible) }$

7. Find the term independent of $x$ in expansion of $\left(\dfrac{\sqrt{x}}{\sqrt{3}}+\dfrac{\sqrt{x}}{2 x^{2}}\right)^{10}$.

Ans: Given, $\left(\dfrac{\sqrt{x}}{\sqrt{3}}+\dfrac{\sqrt{x}}{2 x^{2}}\right)^{10}$.

The term independent of $x$ means there is no term of $x$ in expansion.

Consider $(\mathrm{r}+1)^{\text {th }}$ term be independent of $\mathrm{x}$,

$T_{r+1}={ }^{10} C_{r}\left(\sqrt{\dfrac{x}{3}}\right)^{10-r} \left(\dfrac{\sqrt{3}}{2 x^{2}}\right)^{r}$

$={ }^{10} C_{r}\left(\dfrac{x}{3}\right)^{\dfrac{10-r}{2}} 3^{\dfrac{r}{2}} \cdot \dfrac{1}{2^{r} x^{2 r}}$

$={ }^{10} C_{r} 3^{\dfrac{r}{2}-\dfrac{10-r}{2}} 2^{-r} x^{\dfrac{10-r}{2}-2 r}$

Here, term independent of $\mathrm{x}$,

$\Rightarrow \dfrac{10-\mathrm{r}}{2}-2 \mathrm{r}=0$ $\Rightarrow \mathrm{r}=2$ $\Rightarrow 3^{\mathrm{rd}}$ term is independent of $\mathrm{x}$

$\mathrm{T}_{3}={ }^{10} \mathrm{C}_{2} \dfrac{(3)^{-3}}{4}$ $=\dfrac{10 \times 9}{2 \times 1} \times \dfrac{1}{9 \times 12}$ $=\dfrac{5}{12}$

8. Find the middle term in expansion of $\left(2 a x-\dfrac{b}{x^{2}}\right)^{12}$.

Ans: Given, $\left(2 \mathrm{ax}-\dfrac{\mathrm{b}}{\mathrm{x}^{2}}\right)^{12}$

For even power of binomial, there exist only one middle term. The middle term is given by $\left(\dfrac{\mathrm{n}+2}{2}\right)$.

The middle term is $\left(\dfrac{\mathrm{n}+2}{2}\right)=\dfrac{12+2}{2}=7^{\text {th }}$ term.

$\mathrm{T}_{7}={ }^{12} \mathrm{C}_{6}(2 \mathrm{ax})^{6}\left(-\dfrac{\mathrm{b}}{\mathrm{x}^{2}}\right)^{6}$

$={ }^{12} \mathrm{C}_{6} \dfrac{2^{6} \mathrm{a}^{6} \mathrm{x}^{6}(-\mathrm{b})^{6}}{\mathrm{x}^{12}}$

$={ }^{12} \mathrm{C}_{6} \dfrac{2^{6} \mathrm{a}^{6} \mathrm{~b}^{6}}{\mathrm{x}^{6}}$

$=\dfrac{59136 \mathrm{a}^{6} \mathrm{~b}^{6}}{\mathrm{x}^{6}}$

9. Find the middle term (terms) in the expansion of $\left(\dfrac{\mathbf{p}}{\mathbf{x}}+\dfrac{\mathbf{x}}{\mathbf{p}}\right)^{9}$.

Ans: Given, $\left(\dfrac{\mathrm{p}}{\mathrm{x}}+\dfrac{\mathrm{x}}{\mathrm{p}}\right)^{9}$

For odd power of binomial, there exist two middle terms $\left(\dfrac{\mathrm{n}+1}{2}\right)$ and $\left[\left(\dfrac{\mathrm{n}+1}{2}\right)+1\right]$.

Here, middle term in the expansion are $5^{\text {th }}$ and $6^{\text {th }}$ term.

$\mathrm{T}_{5}={ }^{9} \mathrm{C}_{4}\left(\dfrac{\mathrm{p}}{\mathrm{x}}\right)^{5}\left(\dfrac{\mathrm{x}}{\mathrm{p}}\right)^{4}$

$={ }^{9} \mathrm{C}_{4} \dfrac{\mathrm{p}}{\mathrm{x}}$

$=\dfrac{126 \mathrm{p}}{\mathrm{x}}$

$\mathrm{T}_{6}={ }^{9} \mathrm{C}_{5}\left(\dfrac{\mathrm{p}}{\mathrm{x}}\right)^{4}\left(\dfrac{\mathrm{x}}{\mathrm{p}}\right)^{5}$

$={ }^{9} \mathrm{C}_{4} \dfrac{\mathrm{x}}{\mathrm{p}}$

$=\dfrac{126 \mathrm{x}}{\mathrm{p}}$

10. Show that $2^{4 n+4}-15 n-16$, where $n \in N$ is divisible by 225 .

Ans: Given, $2^{4 n+4}-15 n-16$, where $n \in N$ is divisible by 225 .

Try to find out the factor in $2^{4 n+4}-15 n-16$ which is divisible by 225 .

$2^{4 n+4}-15 n-16=2^{4(n+1)}-15 n-16$

$=16^{n+1}-15 n-16=(15+1)^{n+1}-15 n-16$

$={ }^{n+1} C_{0} 15^{0}+{ }^{n+1} C_{1} 15^{1}+{ }^{n+1} C_{2} 15^{2}+{ }^{n+1} C_{3} 15^{3}+\ldots . .+{ }^{n+1} C_{n+1} 15^{n+1}-15 n-16$

$=1+(n+1) 15+{ }^{n+1} C_{2} 15^{2}+{ }^{n+1} C_{3} 15^{3}+\ldots . .+{ }^{n+1} C_{n+1} 15^{n+1}-15 n-16$

$=1+15 n+15+{ }^{n+1} C_{2} 15^{2}+{ }^{n+1} C_{3} 15^{3}+\ldots . .+{ }^{n+1} C_{n+1} 15^{n+1}-15 n-16$

$=15^{2}\left[{ }^{n+1} C_{2} 15^{2}+{ }^{n+1} C_{3} 15^{3}+\ldots . .+{ }^{n+1} C_{n+1} 15^{n+1}\right]$

$\therefore 2^{4 n+4}-15 n-16 \text { is divisible by } 225$ .

11. Find numerically the greatest term in the expansion of $(2+3 x)^{9}$, where $x=\dfrac{3}{2}$

Ans: Given, $(2+3 x)^{9}$, where $x=\dfrac{3}{2}$

Use general term to find maximum value of $r$ to obtain the numerically greatest term.

$\text { Here, }(2+3 x)^{9}=2^{9}\left(1+\dfrac{3 x}{2}\right)^{9}$

$\dfrac{T_{r+1}}{T_{r}}=\dfrac{2^{9}\left[{ }^{9} C_{r}\left(\dfrac{3 x}{2}\right)^{r}\right]}{2^{9}\left[{ }^{9} C_{r-1}\left(\dfrac{3 x}{2}\right)^{r-1}\right]}$

$=\dfrac{{ }^{9} C_{r}}{{ }^{9} C_{r-1}}\left[\dfrac{3 x}{2}\right].$

$=\dfrac{10-r}{r}\left(\dfrac{9}{4}\right) \quad\left(\text { Here, } x=\dfrac{3}{2}\right)$

$\text { now, } \dfrac{\mathrm{T}_{\mathrm{r}+1}}{\mathrm{~T}_{\mathrm{r}}} \geq 1$

$\Rightarrow \dfrac{90-9 \mathrm{r}}{4 \mathrm{r}} \geq 1$

$\Rightarrow 90-9 \mathrm{r} \geq 4 \mathrm{r}$

$\Rightarrow \mathrm{r} \leq \dfrac{90}{13}$

$\therefore$ Maximum value of $\mathrm{r}=6$

$\Rightarrow \mathrm{T}_{\mathrm{r}+1}=\mathrm{T}_{7}=2^{9}\left[{ }^{9} \mathrm{C}_{6}\left(\dfrac{3 \mathrm{x}}{2}\right)^{6}\right]$

$=2^{9}\cdot { }^{9}\mathrm{C}_{6}\left(\dfrac{3 \mathrm{x}}{2}\right)^{6}$

$=2^{9} \cdot \dfrac{9 \times 8 \times 7}{3 \times 2 \times 1}\left(\dfrac{3^{12}}{2^{12}}\right)$

$=\dfrac{7 \times 3^{13}}{2}$

12. If $\mathrm{n}$ is a positive integer, find coefficient of $\mathrm{x}^{-1}$ in the expansion of $(1+x)^{n}\left(1+\dfrac{1}{x}\right)^{n}$

Ans: Given, $(1+\mathrm{x})^{n}\left(1+\dfrac{1}{\mathrm{x}}\right)^{\mathrm{n}}$.

Simplify the expression. Then, find the coefficient in simplified expression.

$(1+x)^{n}\left(1+\dfrac{1}{x}\right)^{n}=\dfrac{(1+x)^{2 n}}{x^{n}}$

coefficient of $x^{-1}$ in expansion of $\dfrac{(1+x)^{2 n}}{x^{n}}=$ coefficient of $x^{n-1}$ in expansion of $(1+x)^{2 n}$

$(1+x)^{2 n}={ }^{2 n} C_{0} x^{0}+{ }^{2 n} C_{1} x^{1}+{ }^{2 n} C_{2} x^{2}+\ldots \ldots .+{ }^{2 n} C_{n-1} x^{n-1} \ldots .+{ }^{2 n} C_{2 n} x^{2 n}$

$\therefore \text { coefficient of } x^{n-1} ={ }^{2 n} C_{n-1}$

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

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

13. Which of the following is larger?

$99^{50}+100^{50} \text { or } 101^{50} .$

Ans: Given, Values $99^{50}+100^{50} \text { and } 101^{50} .$

$\text { Use } 101^{50}=(100+1)^{50} \text { and } 99^{50}=(100-1)^{50} . \text { Then, compare. }$

$101^{50}=(100+1)^{50}$

$=100^{50}+50(100)^{49}+\dfrac{50.49}{2.1}(100)^{48}+\dfrac{50.49 .48}{3.2 .1}(100)^{47}+\ldots \ldots .$

$\text { similarly, }$

$99^{50}=(100-1)^{50}$

$=100^{50}-50.100^{49}+\dfrac{50.49}{2.1}(100)^{48}-\dfrac{50.49 .48}{3.2 .1}(100)^{47}+\ldots . .$

$\text { Subtract }(2) \text { from }(1),$

$101^{50}-99^{50}=100^{50}+2 . \dfrac{50.49 .48}{3.2 .1} 100^{47}+\ldots . .$

$\Rightarrow 101^{50}-99^{50}>100^{50}$

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

14. Find the coefficient of $x^{50}$ after simplifying and collecting like terms in the expansion $(1+x)^{1000}+x(1+x)^{999}+x^{2}(1+x)^{908}+\ldots \ldots+x^{1000} .$

Ans: Given, $(1+x)^{1000}+x(1+x)^{999}+x^{2}(1+x)^{908}+\ldots \ldots .+x^{1000} .$

The series geometric series with the common ratio $\dfrac{x}{1+x}$. The sum is given by $\mathrm{S}=\dfrac{\mathrm{a}^{\mathrm{n}}(1-\mathrm{r})^{\mathrm{n}+1}}{(1-\mathrm{r})} . .$

$\mathrm{S}=\dfrac{(1+\mathrm{x})^{1000}\left[1-\left(\dfrac{\mathrm{x}}{1+\mathrm{x}}\right)^{1001}\right]}{\left(1-\dfrac{\mathrm{x}}{1+\mathrm{x}}\right)}$

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

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

$\therefore \text { Coefficient of } \mathrm{x}^{50}={ }^{1001} \mathrm{C}_{50}$

$=\dfrac{(1001) !}{50 ! 951 !}$

15. If $a_{1}, a_{2}, a_{3}$ and $a_{4}$ are the coefficient of any four consecutive terms in the expansion of $(1+x)^{n}$, prove that $\dfrac{a_{1}}{a_{1}+a_{2}}+\dfrac{a_{3}}{a_{3}+a_{4}}=\dfrac{2 a_{2}}{a_{2}+a_{3}}$.

Ans: Given, $a_{1}, a_{2}, a_{3}$ and $a_{4}$ are coefficients.

Consider the four consecutive terms as $T_{\mathrm{r}+1}, T_{\mathrm{r}+2}, T_{\mathrm{r}+3}$ and $T_{\mathrm{r}+4}$ with coefficients $a_{1}, a_{2}, a_{3}$ and $a_{4}$ respectively.

$a_{1}=$ coefficient of $T_{r+1}={ }^{n} C_{r}$

$a_{2}=$ coefficient of $T_{r+2}={ }^{n} C_{r+1}$

$a_{3}=$ coefficient of $T_{r+3}={ }^{n} C_{r+2}$

$a_{4}=$ coefficient of $T_{r+4}={ }^{n} C_{r+3}$

now,

$\dfrac{a_{1}}{a_{1}+a_{2}}=\dfrac{{ }^{n} C_{r}}{{ }^{n} C_{r}+{ }^{n} C_{r+1}}$

$=\dfrac{{ }^{n} C_{r}}{{ }^{n+1} C_{r+1}}\left(\because{ }^{n} C_{r}+{ }^{n} C_{r+1}={ }^{n+1} C_{r+1}\right)$

$=\dfrac{n !}{r !(n-r) !} \times \dfrac{(r+1) !(n-r) !}{(n+1) !}$

$=\dfrac{r+1}{n+1}$

$\dfrac{a_{3}}{a_{3}+a_{4}}=\dfrac{{ }^{n} C_{r+2}}{{ }^{n} C_{r+2}+{ }^{n} C_{r+3}}$

$=\dfrac{{ }^{n} C_{r+2}}{{ }^{n+1} C_{r+3}}$

$=\dfrac{r+3}{n+1}$

LHS $=\dfrac{r+1}{n+1}+\dfrac{r+3}{n+1}$

$=\dfrac{2 r+4}{n+1}$

$\text { R.H.S }=\dfrac{2 \mathrm{a}_{2}}{\mathrm{a}_{2}+\mathrm{a}_{3}}$

$=\dfrac{2\left({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}\right)}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+2}}$

$=\dfrac{2\left({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}\right)}{{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}+2}}$

$=2 \cdot \dfrac{\mathrm{n} !}{(\mathrm{r}+1) !(\mathrm{n}-\mathrm{r}+1) !} \times \dfrac{(\mathrm{r}+2) !(\mathrm{n}-\mathrm{r}+1) !}{(\mathrm{n}+1) !}$

$=\dfrac{2 \mathrm{r}+4}{\mathrm{n}+1}$

$\therefore \text { L.H.S }=\text { R.H.S }$

Objective Type Question:

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

(A) 102

(B) 25

(C) 26

(D) None of these

Ans: Correct answer is option (C)

Given, $(x+a)^{51}-(x-a)^{51}$.

Number of terms in the expansion of $(x+a)^{n}$ is $(n+1)$.

The total number of the terms in the expansion of $(x+a)^{51}-(x-a)^{51}$ are 52. Out of 52 terms 26 terms are cancelled.

Therefore, the total number of terms is $52-26=26$.

17. If coefficients of $\mathrm{x}^{7}$ and $\mathrm{x}^{8}$ in $\left(2+\dfrac{\mathrm{x}}{3}\right)^{\mathrm{n}}$ are equal, then $\mathrm{n}$ is

(A) 56

(B) 55

(C) 45

(D) 15

Ans: Correct answer is option (B)

Given, Coefficient of $x^{7}$ and $x^{8}$ are equal.

Use the general term of expansion of $(a+x)^{n}$ is $T_{r+1}={ }^{n} C_{r} a^{n-r} x^{r}$.

$\mathrm{T}_{7+1}={ }^{\mathrm{n}} \mathrm{C}_{7}(2)^{\mathrm{n}-7} \mathrm{x}^{7}$

$\mathrm{~T}_{8}={ }^{\mathrm{n}} \mathrm{C}_{7} \dfrac{2^{\mathrm{n}-7}}{3^{7}} \mathrm{x}^{7}$

and, $\mathrm{T}_{8+1}=\mathrm{T}_{9}$

$\mathrm{T}_{9}={ }^{\mathrm{n}} \mathrm{C}_{8}(2)^{\mathrm{n}-8}\left(\dfrac{\mathrm{x}}{3}\right)^{8}$

$={ }^{\mathrm{n}} \mathrm{C}_{8} \dfrac{(2)^{\mathrm{n}-8}}{3^{8}} \mathrm{x}^{8}$

Since, coefficients are equal.

${ }^{n} C_{7} \dfrac{2^{n-7}}{3^{7}}={ }^{n} C_{8} \dfrac{(2)^{n-8}}{3^{8}}$

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

$\dfrac{8}{n-7}=\dfrac{1}{6}$

$\Rightarrow n=55$

18. If $\left(1-x+x^{2}\right)^{n}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots \ldots . .+a_{2 n} x^{2 n}$, then $a_{0}+a_{2}+a_{4}+\ldots .+a_{2 n}$ equals.

(A) $\dfrac{3^{n}+1}{2}$

(C) $\dfrac{3^{n}-1}{2}$

(C) $\dfrac{1-3^{n}}{2}$

(D) $3^{\mathrm{n}}+\dfrac{1}{2}$

Ans: Correct answer is option (A)

Given, Expansion of $\left(1-x+x^{2}\right)^{n}$.

Substitute the values in expansion of $\left(1-x+x^{2}\right)^{n}$. Then, obtain a sum of even coefficients.

$\left(1-x+x^{2}\right)^{n}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots . .+a_{2 n} x^{2 n}$

for $x=1$

$1=a_{0}+a_{1}+a_{2}+a_{3}+\ldots \ldots . a_{2 n}-(1)$

for $x=-1$

$3^{n}=a_{0}-a_{1}+a_{2}-a_{3}+\ldots \ldots .+a_{2 n}-(2)$

from (1) and (2),

$3^{n}+1=2\left(a_{0}+a_{2}+a_{4}+\ldots .+a_{2 n}\right)$

$\therefore a_{0}+a_{2}+a_{4}+\ldots .+a_{2 n}=\dfrac{3^{n}+1}{2}$

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

(A) Equal

(B) Equal with opposite signs

(C) Reciprocal of each other

(D) None of these

Ans: Correct answer is option (A)

Given, $(1+x)^{p+q}$

Strategy: Coefficient of $x^{p}$ and $x^{q}$ in the expansion of $(1+x)^{p+q}$ are

${ }^{p+q} C_{p}=\dfrac{(p+q) !}{p ! q !}$

${ }^{p+q} C_{p} \text { and }{ }^{p+q} C_{q} \cdot{ }^{p+q} C_{q}=\dfrac{(p+q) !}{p ! q !}$

$\therefore{ }^{p+q} C_{p}={ }^{p+q} C_{q}$

20. The number of terms in the expansion of $(a+b+c)^{n}$, where $n \in N$ is

(A) $\dfrac{(n+1)(n+2)}{2}$

(B) $n+1$

(C) $n+2$

(D) $(n+1) n$

Ans: Correct answer is option (A)

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

Expand $(a+b+c)^{n}$. Then, analyze the number of terms in the form of a series.

$(a+b+c)^{n}=[a+(b+c)]^{n}$

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

Here, $\mathrm{r}^{\text {th }}$ term contains $\mathrm{r}$ number of terms.

$\therefore \text { Total number of terms }=1+2+3+\ldots \ldots+(n+1)$

$=\dfrac{(n+1)(n+2)}{2}$

21. The ratio of coefficient of $x^{15}$ to the term independent of $x$ in expansion of $\left(x^{2}+\dfrac{2}{x}\right)^{15}$ is

(a) $12: 32$

(b) $1: 32$

(c) $32: 12$

(d) $32: 1$

Ans: Correct answer is option (B)

Given, $\left(x^{2}+\dfrac{2}{x}\right)^{15}$

Consider the general term $T_{\mathrm{r}+1}$. Then, find the term which does not contain $x$.

$\mathrm{T}_{\mathrm{r}+1}={ }^{15} \mathrm{C}_{\mathrm{r}}\left(\mathrm{x}^{2}\right)^{15-\mathrm{r}}\left(\dfrac{2}{\mathrm{x}}\right)^{\mathrm{r}}$

$={ }^{15} \mathrm{C}_{\mathrm{r}}(2)^{\mathrm{r}} \mathrm{x}^{30-3 \mathrm{r}}$

$\text { for } \mathrm{x}^{15}, 30-3 \mathrm{r}=15$

$\Rightarrow \mathrm{r}=5$

$\therefore \text { Coefficient of } \mathrm{x}^{15}={ }^{15} \mathrm{C}_{5}(2)^{5}$

$\text { For } x^{0}, 30-3 r=0$

$\Rightarrow r=10$

$\therefore \text { Coefficient of } x^{0}={ }^{15} C_{10}(2)^{10}$

$\therefore \dfrac{x^{15}}{x^{0}}=\dfrac{{ }^{15} C_{5}(2)^{5}}{{ }^{15} C_{10}(2)^{10}}$

$=\dfrac{1}{2^{5}}$

$=\dfrac{1}{32}$

22. If $z=\left(\dfrac{\sqrt{3}}{2}+\dfrac{i}{2}\right)^{5}+\left(\dfrac{\sqrt{3}}{2}-\dfrac{i}{2}\right)^{5}$, then

(A) $\operatorname{Re}(\mathbf{z})=0$

(B) $I_{\mathrm{m}}(z)=0$

(C) $\operatorname{Re}(\mathbf{z})>\mathbf{0}, \mathbf{I}_{\mathrm{m}}(\mathbf{z})>\mathbf{0}$

(D) $\operatorname{Re}(\mathrm{z})>0, \mathrm{I}_{\mathrm{m}}(\mathrm{z})<0$

Ans: Correct answer is option (B)

Given, $z=\left(\dfrac{\sqrt{3}}{2}+\dfrac{i}{2}\right)^{5}+\left(\dfrac{\sqrt{3}}{2}-\dfrac{i}{2}\right)^{5}$

$\operatorname{Re}(z)$ and $I_{m}(z)$ are real and imaginary part. Use $i^{2}=-1$ and $i^{2}=4$

$\mathrm{z}=2\left[{ }^{5} \mathrm{C}_{0}\left(\dfrac{\sqrt{3}}{2}\right)^{2}+{ }^{5} \mathrm{C}_{2}\left(\dfrac{\sqrt{3}}{2}\right)^{3}\left(\dfrac{\mathrm{i}}{2}\right)^{3}+{ }^{5} \mathrm{C}_{4}\left(\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{\mathrm{i}}{2}\right)^{4}\right]$

here, $i^{2}=-1$ and $i^{2}=4$.

There is no imaginary term. Therefore, $\mathrm{I}_{\mathrm{m}}(\mathrm{z})=0$.

EXERCISE 8.1

1. Find the term independent of $\mathrm{x}$,  $x \neq 0$, in the expansion of $\left(\dfrac{3 x^{2}}{2}-\dfrac{1}{3 x}\right)^{15}$.

Ans: Given, $\left(\dfrac{3 x^{2}}{2}-\dfrac{1}{3 x}\right)^{15}$

Use the general term in expansion of $(x-a)^{n}$ is $T_{r+1}={ }^{n} C_{r}(x)^{n-r}(-a)^{r}$. The term independent of $x$ is obtained by substituting $n-r=0$.

$\mathrm{T}_{\mathrm{r}+1}={ }^{15} \mathrm{C}_{\mathrm{r}}\left(\dfrac{3 \mathrm{x}^{2}}{2}\right)^{15-\mathrm{r}}\left(-\dfrac{1}{3 \mathrm{x}}\right)^{\mathrm{r}}$

$={ }^{15} \mathrm{C}_{\mathrm{r}} 3^{15-\mathrm{r}} \mathrm{x}^{30-2 \mathrm{r}} 2^{\mathrm{r}-15}(-1)^{\mathrm{r}} 3^{-\mathrm{r}} \mathrm{x}^{-\mathrm{r}}$

$={ }^{15} \mathrm{C}_{\mathrm{r}}(-1)^{\mathrm{r}} 3^{15-2 \mathrm{r}} 2^{\mathrm{r}-15} \mathrm{x}^{30-3 \mathrm{r}}$

for independent of $x, 30-3 r=0$

$\Rightarrow \mathrm{r}=10$

$\therefore \mathrm{T}_{\mathrm{r}+1}={ }^{15} \mathrm{C}_{10}(-1)^{10} 3^{15-20} 2^{10-15}$

$={ }^{15} \mathrm{C}_{10} 3^{-5} 2^{-5}$

$={ }^{15} \mathrm{C}_{10} 6^{-5}$

$={ }^{15} \mathrm{C}_{10}\left(\dfrac{1}{6}\right)^{5}$

2. If the term free from $x$ in the expansion of $\left(\sqrt{x}-\dfrac{k}{x^{2}}\right)^{10}$ is 405 , then find value of $\mathrm{k}$.

Ans: Given, Term independent of $\mathrm{x}$ is 405 .

Use the general term in expansion of $(x-a)^{n}$ is $T_{r+1}={ }^{n} C_{r}(x)^{n-r}(-a)^{r}$.

The term independent of $x$ is obtained by substituting $n-r=0$.

$\mathrm{T}_{\mathrm{r}+1}={ }^{10} \mathrm{C}_{\mathrm{r}}(\sqrt{\mathrm{x}})^{10-\mathrm{r}}\left(-\dfrac{\mathrm{k}}{\mathrm{x}^{2}}\right)^{\mathrm{r}}$

$={ }^{10} \mathrm{C}_{\mathrm{r}}(\mathrm{x})^{\dfrac{1}{2}(10-\mathrm{r})}(-\mathrm{k})^{\mathrm{r}} \mathrm{x}^{-2 \mathrm{r}}$

$={ }^{10} \mathrm{C}_{\mathrm{r}}(\mathrm{x})^{5-\dfrac{r}{2}-2 \mathrm{r}}(-\mathrm{k})^{\mathrm{r}}$

$={ }^{10} \mathrm{C}_{\mathrm{r}}(\mathrm{x})^{\dfrac{10-5 \mathrm{r}}{2}}(-\mathrm{k})^{\mathrm{r}}$

For independent of $x, \dfrac{10-5 r}{2}=0$

$\Rightarrow \mathrm{r}=2$

$\therefore \mathrm{T}_{2+1}={ }^{10} \mathrm{C}_{\mathrm{r}}(-\mathrm{k})^{2}=405$

$\Rightarrow \dfrac{10 \times 9 \times 8 !}{2 ! \times 8 !}{k}^{2}=405$

$\Rightarrow 45 \mathrm{k}^{2}=405$

$\Rightarrow \mathrm{k}^{2}=\dfrac{405}{45}$

$\Rightarrow \mathrm{k}^{2}=9$

$\therefore \mathrm{k}=\pm 3$

3. Find the coefficient of $x$ in the expansion of $\left(1-3 x+7 x^{2}\right)(1-x)^{16}$.

Ans: Given, $\left(1-3 x+7 x^{2}\right)(1-x)^{16}$.

Expand the expansion of $(1-\mathrm{x})^{16}$.

Then, find coefficient of $\mathrm{x}$ in the expansion of $\left(1-3 x+7 x^{2}\right)(1-x)^{16}$.

$\left(1-3 x+7 x^{2}\right)(1-x)^{16}=\left(1-3 x+7 x^{2}\right)\left({ }^{16} C_{0} 1^{16}-{ }^{16} C_{1} 1^{15} x^{1}+{ }^{16} C_{2} 1^{14} x^{2}+\ldots \ldots+{ }^{16} C_{16} x^{16}\right)$

$=\left(1-3 x+7 x^{2}\right)\left(1-16 x+120 x^{2}+\ldots . .\right)$

$\therefore$ Coefficient of $x=-3-16=-19$

4. Find the term independent of $x$ in the expansion of $\left(3 x-\dfrac{2}{x^{15}}\right)^{15}$.

Ans: Given, $\left(3 x-\dfrac{2}{x^{15}}\right)^{15}$

Use the general term in expansion of $(x-a)^{n}$ is $T_{r+1}={ }^{n} C_{r}(x)^{n-r}(-a)^{r}$.

The term independent of $x$ is obtained by substituting $n-r=0$.

$\mathrm{T}_{\mathrm{r}+1}={ }^{15} \mathrm{C}_{\mathrm{r}}(3 \mathrm{x})^{15-\mathrm{r}}\left(-\dfrac{2}{\mathrm{x}^{2}}\right)^{\mathrm{r}}$

$={ }^{15} \mathrm{C}_{\mathrm{r}}(3 \mathrm{x})^{15-\mathrm{r}}(-2)^{\mathrm{r}} \mathrm{x}^{-2 \mathrm{r}}$

$={ }^{15} \mathrm{C}_{\mathrm{r}} 3^{15-\mathrm{r}} \mathrm{x}^{15-3 \mathrm{r}}(-2)^{\mathrm{r}}$

for independent of $\mathrm{x}$,

$15-3 r=0$

$\Rightarrow r=5$

$T_{5+1}={ }^{15} C_{5} 3^{15-5}(-2)^{5}$

$T_{6}=-\dfrac{15 \times 14 \times 13 \times 12 \times 11 \times 10 !}{5 \times 4 \times 3 \times 2 \times 1 \times 10 !} .3^{10} .2^{5}$

$T_{6}=-3003.3^{10} .2^{5}$

5. Find the middle term (terms) in the expansion of

(i) $\left(\dfrac{x}{a}-\dfrac{a}{x}\right)^{10}$

Ans: Given, $\left(\dfrac{x}{a}-\dfrac{a}{x}\right)^{10}$

In even power of binomial there exists one middle term $\left(\dfrac{\mathrm{n}}{2}+1\right)^{th}$ terms. For odd power of binomial there exists two middle terms $\left(\dfrac{\mathrm{n}+1}{2}\right)^{th}$ and $\left(\dfrac{\mathrm{n}+1}{2}+1\right)^{th}$ terms.

Here,

$\mathrm{n}=10$

$\Rightarrow$ Middle term is $\left(\dfrac{10}{2}+1\right)^{th}$ $=6^{th}$ term.

$\therefore \mathrm{T}{5+1}=\mathrm{T}{6}={ }^{10} \mathrm{C}_{5}\left(\dfrac{\mathrm{x}}{\mathrm{a}}\right)^{5}\left(-\dfrac{\mathrm{a}}{\mathrm{x}}\right)^{5}$

$=-{ }^{10} \mathrm{C}_{5}\left(\dfrac{\mathrm{x}}{\mathrm{a}}\right)^{5}\left(\dfrac{\mathrm{a}}{\mathrm{x}}\right)^{5}$

$=-\dfrac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 \times 4 \times 3 \times 2 \times 1}$

$=-252$

(ii) $\left(3 x-\dfrac{x^{3}}{6}\right)^{9}$

Ans: Given, $\left(3 x-\dfrac{x^{3}}{6}\right)^{9}$

In even power of binomial there exists one middle term $\left(\dfrac{\mathrm{n}}{2}+1\right)^{th}$ terms. For odd power of binomial there exists two middle terms $\left(\dfrac{n+1}{2}\right)^{th}$ and $\left(\dfrac{n+1}{2}+1\right)^{th}$ terms.

Here,

$\mathrm{n}=9 \text {, }$

$\Rightarrow \text { Middle terms are }\left(\dfrac{9+1}{2}\right)=5^{\text {th }} \text { term and }\left(\dfrac{9+1}{2}+1\right)^{ th }=6 ^\text { th}$ term

$\therefore \mathrm{T}_{4+1}=\mathrm{T}_{5}={ }^{9} \mathrm{C}_{4}(3 \mathrm{x})^{9-4}\left(-\dfrac{\mathrm{x}^{3}}{6}\right)^{4}$

$=\dfrac{9 \times 8 \times 7 \times 6 \times 5 !}{4 \times 3 \times 2 \times 1 \times 5 !} \cdot 3^{5} \mathrm{x}^{5} \mathrm{x}^{12} 6^{-4}$

$=\dfrac{7 \times 6 \times 3 \times 3^{1}}{2^{4}} \mathrm{x}^{17}$

$=\dfrac{189}{8} \mathrm{x}^{17}$

$\therefore \mathrm{T}_{5+1}=\mathrm{T}_{6}={ }^{9} \mathrm{C}_{4}(3 \mathrm{x})^{9-5}\left(-\dfrac{\mathrm{x}^{3}}{6}\right)^{5}$

$=-\dfrac{9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 4 \times 3 \times 2 \times 1} \cdot 3^{4} \mathrm{x}^{4} \mathrm{x}^{15} 6^{-5}$

$=\dfrac{-21 \times 6}{3 \times 2^{5}} \mathrm{x}^{19}$

$=\dfrac{-21}{16} \mathrm{x}^{19}$

6. Find the coefficient of $x^{15}$ in the expansion of $\left(x-x^{2}\right)^{10}$.

Ans: Given, $\left(x-x^{2}\right)^{10}$

Use the general term in expansion of $(x-a)^{n}$ is $T_{r+1}={ }^{n} C_{r}(x)^{n-r}(-a)^{r}$.

$\mathrm{T}_{\mathrm{r}+1}={ }^{10} \mathrm{C}_{\mathrm{r}}(\mathrm{x})^{10-\mathrm{r}}\left(-\mathrm{x}^{2}\right)^{\mathrm{r}}$

$=(-1)^{\mathrm{r}} \cdot{ }^{10} \mathrm{C}_{\mathrm{r}}(\mathrm{x})^{10-\mathrm{r}} \cdot \mathrm{x}^{2 \mathrm{r}}$

$=(-1)^{\mathrm{r}} \cdot{ }^{10} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{10+\mathrm{r}}$

For coefficient is $x^{15}$,

$10+r=15$

$\Rightarrow r=5$

$T_{5+1}=(-1)^{5} \cdot{ }^{10} \mathrm{C}_{5} x^{15}$

$\therefore$ Coefficients of $x^{15}=-\dfrac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 \times 4 \times 3 \times 2 \times 1 \times 5 !}$

$=-3 \times 2 \times 7 \times 6$

$=-252$

7. Find the coefficient of $\dfrac{1}{x^{17}}$ in the expansion of $\left(x^{4}-\dfrac{1}{x^{3}}\right)^{15}$.

Ans: Given, $\left(x^{4}-\dfrac{1}{x^{3}}\right)^{15}$.

Use the general term in expansion of $(x-a)^{n}$ is $T_{r+1}={ }^{n} C_{r}(x)^{n-r}(-a)^{r}$.

$T_{r+1}={ }^{n} C_{r}(x)^{n-r}(-a)^{r}$

$={ }^{15} C_{r}\left(x^{4}\right)^{15-r}\left(-\dfrac{1}{x^{3}}\right)^{r}$

$={ }^{15} C_{r}(x)^{60-4 r}(-1)^{r} x^{-3 r}$

$={ }^{15} C_{r}(x)^{60-7 r}(-1)^{r}$

For coefficient of $x^{-17}$,

$60-7 r=-17$

$\Rightarrow 7 r=77$

$\Rightarrow r=11$

$\therefore T_{11+1}={ }^{15} C_{11} x^{60-77}(-1)^{11}$

$\text { Coefficient of } \mathrm{x}^{-17}=\dfrac{-15 \times 14 \times 13 \times 12 \times 11 !}{11 ! \times 4 \times 3 \times 2 \times 1}$

$=-15 \times 17 \times 13$

$=-1365$

8. Find the sixth term of the expansion $\left(y^{\dfrac{1}{2}}+x^{\dfrac{1}{3}}\right)^{n}$, if the binomial coefficient of the third term from the end is 45 .

Ans: Given, $\left(\mathrm{y}^{\dfrac{1}{2}}+\mathrm{x}^{\dfrac{1}{3}}\right)^{\mathrm{n}}$

Binomial coefficient of the third term from the end is same as the binomial coefficient of third term form beginning. Use ${ }^{n} \mathrm{C}_{2}=45$.

$\mathrm{T}_{5+1}=\mathrm{T}_{6}={ }^{\mathrm{n}} \mathrm{C}_{5}\left(\mathrm{y}^{\dfrac{1}{2}}\right)^{\mathrm{n}-5}\left(\mathrm{x}^{\dfrac{1}{3}}\right)^{5}$

now, ${ }^{n} \mathrm{C}_{2}=45$

$\Rightarrow \dfrac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2) !}{2 !(\mathrm{n}-2) !}=45$ $\Rightarrow \mathrm{n}(\mathrm{n}-1)=90$ $\Rightarrow \mathrm{n}^{2}-\mathrm{n}-90=0$

$\Rightarrow \mathrm{n}^{2}-10 \mathrm{n}-\mathrm{n}-90=0$

$\Rightarrow(\mathrm{n}-10)(\mathrm{n}+9)=0$

$\Rightarrow \mathrm{n}=-9$ (not possible) or $\mathrm{n}=10$

$\therefore \mathrm{T}_{6}={ }^{10} \mathrm{C}_{5}\left(\mathrm{y}^{\dfrac{1}{2}}\right)^{10-5}\left(\mathrm{x}^{\dfrac{1}{3}}\right)^{5}$

$=252 \mathrm{y}^{\dfrac{5}{2}} \cdot \mathrm{x}^{\dfrac{5}{3}}$

9. Find the value of $r$, if the coefficients of $(2 r+4)^{\text {th }}$ and $(r-2)^{\text {th }}$ terms in the expansion of $(1+x)^{18}$ are equal.

Ans: Given, Coefficients of $(2 r+4)^{\text {th }}$ and $(r-2)^{\text {th }}$ terms are equal.

Coefficient of $(r+1)^{\text {th }}$ term in the expansion of $(1+x)^{n}$ is ${ }^{n} C_{r}$.

$(2 \mathrm{r}+4)^{\mathrm{th}} \text { term is } \mathrm{T}_{2 \mathrm{r}+3+1}$

$\mathrm{~T}_{2 \mathrm{r}+4 \mathrm{r}}={ }^{18} \mathrm{C}_{2 \mathrm{r}+3}(1)^{18-2 \mathrm{r}-3}(\mathrm{x})^{2 \mathrm{r}+3}$

$={ }^{18} \mathrm{C}_{2 \mathrm{r}+3}(\mathrm{x})^{2 \mathrm{r}+3}$

$({r}-2)^{\mathrm{th}} \text { term is } \mathrm{T}_{\mathrm{r}-3+1} .$

$\mathrm{T}_{\mathrm{r}-2}={ }^{18} \mathrm{C}_{\mathrm{r}-3} \mathrm{x}^{\mathrm{r}-3}$

$\text { Here, }{ }^{18} \mathrm{C}_{2 \mathrm{r}+3}={ }^{18} \mathrm{C}_{\mathrm{r}-3}$

$\Rightarrow 2 \mathrm{r}+3=\mathrm{r}-3$

$\Rightarrow \mathrm{r}=6$

10. If the coefficient of second, third and fourth terms in the expansion of $(1+x)^{2 n}$ are in AP, then show that $2 n^{2}-9 n+7=0$.

Ans: Given, Coefficients of second, third and fourth are in AP.

In expansion of $(x+y)^{n}$, the coefficient of $(r+1)^{\text {th }}$ term is ${ }^{n} C_{r}$. If $a$, $b$ and $c$ are in AP, then $2 \mathrm{~b}=\mathrm{a}+\mathrm{c}$.

In $(1+x)^{2 n}$,

coefficient of second term $={ }^{2 n} C_{1}$

coefficient of third term $={ }^{2 \mathrm{n}} C_{2}$

coefficient of fourth term $={ }^{2 n} C_{3}$

Since, coefficients are in AP.

$\therefore 2^{2 n} C_{2}={ }^{2 n} C_{1}+{ }^{2 n} C_{3}$

$\Rightarrow 2\left[\dfrac{2 n(2 n-1)(2 n-2) !}{2 \times 1 \times(2 n-2) !}\right]=\dfrac{2 n(2 n-1) !}{(2 n-1) !}+\dfrac{2 n(2 n-1)(2 n-2)(2 n-3) !}{3 !(2 n-3) !}$

$\Rightarrow n(2 n-1)=n+\dfrac{n(2 n-1)(2 n-2)}{6}$

$\Rightarrow n(12 n-6)=n\left(6+4 n^{2}-4 n-2 n+2\right)$

$\Rightarrow 6(2 n-1)=2\left(2 n^{2}-3 n+4\right)$

$\Rightarrow 3(2 n-1)=2 n^{2}-3 n+4$

$\Rightarrow 2 n^{2}-3 n+4-6 n+3=0$

$\Rightarrow 2 n^{2}-9 n+7=0$

11. Find the coefficient of $x^{4}$ in the expansion of $\left(1+x+x^{2}+x^{3}\right)^{11}$.

Ans: Given, $\left(1+x+x^{2}+x^{3}\right)^{11}$.

Simplify the expression. Then, expand.

$\left(1+x+x^{2}+x^{3}\right)^{11}=\left[(1+x)+x^{2}(1+x)\right]^{11}$

$=\left[(1+x)\left(1+x^{2}\right)\right]^{11}$

$=(1+x)^{11}\left(1+x^{2}\right)^{11}$

$=\left({ }^{11} C_{0}+{ }^{11} C_{1} x+{ }^{11} C_{2} x^{2}+{ }^{11} C_{3} x^{3}+{ }^{11} C_{4} x^{4}+\ldots\right)\left({ }^{11} C_{0}+{ }^{11} C_{1} x^{2}+{ }^{11} C_{2} x^{4}+\ldots . .\right)$

$=\left(1+11 x+55 x^{2}+165 x^{3}+330 x^{4}+\ldots\right)\left(1+11 x^{2}+55 x^{4}+\ldots . .\right)$

$\therefore \text { Coefficient of } x^{4}=55+605+330=990$

12. If $\mathrm{p}$ is a real number and the middle term in the expansion of $\left(\dfrac{p}{2}+2\right)^{8}$ is 1120 , then find the value of $p$.

Ans: Given, Middle term in expansion of $\left(\dfrac{\mathrm{p}}{2}+2\right)^{8}$ is 1120 .

In even power of binomial there exists one middle term $\left(\dfrac{\mathrm{n}}{2}+1\right)^{th}$ terms.

For, $\mathrm{n}=8$

$\left(\dfrac{8}{2}+1\right)^{th}$ $=5^{th}$ term $\mathrm{T}_{4+1}=\mathrm{T}_{5}={ }^{8} \mathrm{C}_{4}\left(\dfrac{\mathrm{p}}{2}\right)^{8-4} \cdot 2^{4}$

$\Rightarrow 1120={ }^{8} \mathrm{C}_{4} \mathrm{p}^{4} \cdot 2^{-4} \cdot 2^{4}$

$\Rightarrow 1120=\dfrac{8 \times 7 \times 6 \times 5 \times 4 !}{4 ! \times 4 \times 3 \times 2 \times 1} \mathrm{p}^{4}$

$\Rightarrow 1120=7 \times 2 \times 5 \times \mathrm{p}^{4}$

$\Rightarrow \mathrm{p}^{4}=\dfrac{1120}{70}=16$

$\Rightarrow \mathrm{p}^{4}=2^{4}$

$\Rightarrow \mathrm{p}=\pm 2$

13. Show that the middle term in the expansion of $\left(x-\dfrac{1}{x}\right)^{2 n}$ is $\dfrac{1 \times 3 \times 5 \times \ldots . . \times(2 n-1)}{n !} \times(-2)^{n}$

Ans: Given, $\left(x-\dfrac{1}{x}\right)^{2 n}$

Show that the middle term is $\dfrac{1 \times 3 \times 5 \times \ldots . . \times(2 \mathrm{n}-1)}{\mathrm{n} !} \times(-2)^{\mathrm{n}}$.

In even power of binomial there exists one middle term $\left(\dfrac{\mathrm{n}}{2}+1\right)^{th}$ terms.

Middle term is $\left(\dfrac{2 n}{2}+1\right)^{th}$ $=(n+1)^{th}$ term.

$T_{n+1}={ }^{2 n} C_{n}(x)^{2 n-n}\left(-\dfrac{1}{x}\right)^{n}$

$={ }^{2 n} C_{n}(-1)^{n}$

$=(-1)^{n} \dfrac{(2 n) !}{n ! n !}$

$=\dfrac{1.2 .3 .4 .5 \ldots \ldots .(2 n-1)(2 n)}{n ! n !}(-1)^{n}$

$=\dfrac{1.3 .5 \ldots(2 n-1) .2 .4 .6 \ldots(2 n)}{(1.2 .3 \ldots \ldots n)(n !)}(-1)^{n}$

$=\dfrac{1.3 .5 \ldots(2 n-1) .2^{n}(1.2 .3 \ldots n)(-1)^{n}}{(1.2 .3 \ldots \ldots n)(n !)}$

$=\dfrac{[1.3 .5 \ldots(2 n-1)]}{n !}(-2)^{n}$

14. Find $n$ in the binomial $\left(\sqrt[3]{2}+\dfrac{1}{\sqrt[3]{3}}\right)^{n}$, if the ratio of $7^{th }$ term from the beginning to the $7^{th}$ term from the end is $\dfrac{1}{6}$.

Ans: Given, $\left(\sqrt[3]{2}+\dfrac{1}{\sqrt[3]{3}}\right)^{\mathrm{n}}$

Using the general term find the $7^{th}$ term from the beginning to the $7^{th}$ term from the end.

$7^{th}$ term from the beginning $\mathrm{T}_{7}=\mathrm{T}_{6+1}={ }^{\mathrm{n}} \mathrm{C}_{6}(\sqrt[3]{2})^{\mathrm{n}-6}\left(\dfrac{1}{\sqrt[3]{3}}\right)^{6}$

$7^{th}$ term from the end $=\mathrm{T}_{7}$ from beginning of $\left(\dfrac{1}{\sqrt[3]{3}}+\sqrt[3]{2}\right)^{\mathrm{n}}$

$\therefore \mathrm{T}_{7}={ }^{\mathrm{n}}$ $\mathrm{C}_{6}\left(\dfrac{1}{\sqrt[3]{3}}\right)^{\mathrm{n}-6}(\sqrt[3]{2})^{6}$

now,

$\dfrac{{ }^{n} C_{6}(\sqrt[3]{2})^{n-6}\left(\dfrac{1}{\sqrt[3]{3}}\right)^{6}}{{ }^{n} C_{6}\left(\dfrac{1}{\sqrt[3]{3}}\right)^{n-6}(\sqrt[3]{2})^{6}}=\dfrac{1}{6}$

$\Rightarrow \dfrac{2^{\dfrac{n-6}{3}} \cdot 3^{-\dfrac{6}{3}}}{3^{-\left(\dfrac{n-6}{3}\right)} \cdot 2^{\dfrac{6}{3}}}=\dfrac{1}{6}$

$\Rightarrow\left(2^{\dfrac{n-6}{3}} \cdot 3^{-\dfrac{6}{3}}\right)\left(3^{-\left(\dfrac{n-6}{3}\right)} \cdot 2^{\dfrac{6}{3}}\right)=6^{-1}$

$\Rightarrow(2.3)^{\dfrac{n}{3}-4}=6^{-1}$

$\Rightarrow \dfrac{n}{3}-4=-1$

$\Rightarrow \mathrm{n}=9$

15. In the expansion of $(x+a)^{n}$, if the sum of odd number is denoted by $\mathrm{O}$ and the sum of even term by E. Then, prove that

(i) $\mathbf{O}^{2}-\mathbf{E}^{2}=\left(\mathbf{x}^{2}-\mathrm{a}^{2}\right)^{\mathrm{n}}$.

Ans: Given, $(x+a)^{n}$.

Use the general expansion. Find the sum of odd and even terms.

Here,

$(x+a)^{n}={ }^{n} C_{0} x^{n} a^{0}+{ }^{n} C_{1} x^{n-1} a^{1}+{ }^{n} C_{2} x^{n-2} a^{2}+\ldots .+{ }^{n} C_{n} a^{n}$

$\therefore O={ }^{n} C_{0} x^{n}+{ }^{n} C_{2} x^{n-2} a^{2}+\ldots . .$

$E={ }^{n} C_{1} x^{n-1} a^{1}+{ }^{n} C_{3} x^{n-3} a^{3}+\ldots . .$

now,

$(x+a)^{n}=O+E \quad-(1)$

$(x-a)^{n}=O-E-(2)$

$\therefore(O+E)(O-E)=(x+a)^{n}(x-a)^{n}$

$\Rightarrow O^{2}-E^{2}=\left(x^{2}-a^{2}\right)^{2 n}$

(ii) $4 O E=(x+a)^{2 n}-(x-a)^{2 n}$.

Ans: Given, $(x+a)^{n}$.

Use the general expansion. Find the sum of odd and even terms.

Here,

$4 O E=(O+E)^{2}-(O-E)^{2}$

$=\left[(x+a)^{n}\right]^{2}-\left[(x-a)^{n}\right]^{2}$

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

16. If $x^{p}$ occurs in the expansion of $\left(x^{2}+\dfrac{1}{x}\right)^{2 n}$, then prove that its coefficient is $\dfrac{2 n !}{\dfrac{(4 n-p) !}{3 !} \dfrac{(2 n+p) !}{3 !}}$

Ans: Given, $\left(x^{2}+\dfrac{1}{x}\right)^{2 n}$.

Use the general term in expansion of $(x-a)^{n}$ is $T_{r+1}={ }^{n} C_{r}(x)^{n-r}(-a)^{r}$.

$\mathrm{T}_{\mathrm{r}+1}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}}\left(\mathrm{x}^{2}\right)^{2 \mathrm{n}-\mathrm{r}}\left(\dfrac{1}{\mathrm{x}}\right)^{\mathrm{r}}$

$={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{4 \mathrm{n}-2 \mathrm{r}} \mathrm{x}^{-\mathrm{r}}$

$={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{4 \mathrm{n}-3 \mathrm{r}}$

Here, $4 \mathrm{n}-3=\mathrm{p}$

$\Rightarrow \mathrm{r}=\dfrac{4 \mathrm{n}-\mathrm{p}}{3}$

$\therefore \text { Coefficient of } \mathrm{x}^{\mathrm{p}}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}}$

$=\dfrac{(2 \mathrm{n}) !}{\mathrm{r} !(2 \mathrm{n}-\mathrm{r}) !}$

$=\dfrac{(2 \mathrm{n}) !}{\left(\dfrac{4 \mathrm{n}-\mathrm{p}}{3}\right) !\left(2 \mathrm{n}-\dfrac{4 \mathrm{n}-\mathrm{p}}{3}\right) !}$

$=\dfrac{(2 \mathrm{n}) !}{\left(\dfrac{4 \mathrm{n}-\mathrm{p}}{3}\right) !\left(\dfrac{2 \mathrm{n}+\mathrm{p}}{3}\right) !}$

17. Find the term independent of $x$ in the expansion of $\left(1+x+2 x^{3}\right)\left(\dfrac{3}{2} x^{2}-\dfrac{1}{3 x}\right)^{9}$

Ans: Given, $\left(1+x+2 x^{3}\right)\left(\dfrac{3}{2} x^{2}-\dfrac{1}{3 x}\right)^{9}$.

Use the general term in expansion of $(x-a)^{n}$ is $T_{r+1}={ }^{n} C_{r}(x)^{n-r}(-a)^{r}$. The term independent of $x$ is obtained by substituting $n-r=0$.

Consider, $\left(\dfrac{3}{2} \mathrm{x}^{2}-\dfrac{1}{3 \mathrm{x}}\right)^{9}$

$=\mathrm{T}_{\mathrm{r}+1}={ }^{9} \mathrm{C}_{\mathrm{r}}\left(\dfrac{3}{2} \mathrm{x}^{2}\right)^{9-\mathrm{r}}\left(-\dfrac{1}{3 \mathrm{x}}\right)^{\mathrm{r}}$

$={ }^{9} \mathrm{C}_{\mathrm{r}}\left(\dfrac{3}{2}\right)^{9-\mathrm{r}} \mathrm{x}^{18-2 \mathrm{r}}\left(-\dfrac{1}{3}\right)^{\mathrm{r}} \mathrm{x}^{-\mathrm{r}}$

$={ }^{9}\mathrm{C}_{\mathrm{r}}\left(\dfrac{3}{2}\right)^{9-\mathrm{r}}\left(-\dfrac{1}{3}\right)^{\mathrm{r}} \mathrm{x}^{18-3 \mathrm{r}}$

$\therefore$ The general term of expansion $\left(1+x+2 x^{3}\right)\left(\dfrac{3}{2} x^{2}-\dfrac{1}{3 x}\right)^{9}$

$\therefore$ The general term of expansion $\left(1+x+2 x^{3}\right)\left(\dfrac{3}{2} x^{2}-\dfrac{1}{3 x}\right)^{9}$

$={ }^{9}\mathrm{C}_{\mathrm{r}}\left(\dfrac{3}{2}\right)^{9-\mathrm{r}}\left(-\dfrac{1}{3}\right)^{\mathrm{r}} \mathrm{x}^{18-3 \mathrm{r}}+{ }^{9} \mathrm{C}_{\mathrm{r}}\left(\dfrac{3}{2}\right)^{9-\mathrm{r}}\left(-\dfrac{1}{3}\right)^{\mathrm{r}} \mathrm{x}^{19-3 \mathrm{r}}+2{ }^{9} \mathrm{C}_{\mathrm{r}}\left(\dfrac{3}{2}\right)^{9-\mathrm{r}}\left(-\dfrac{1}{3}\right)^{\mathrm{r}} \mathrm{x}^{21-3 \mathrm{r}}$ for term independent of $x$,

$18-3 r=0,19-3 r=0$ and $21-3 r=0$

$\Rightarrow \mathrm{r}=6, \dfrac{19}{3}$ (not possible), 7

$\therefore$ Term independent of $\mathrm{x}$

$={ }^{9} \mathrm{C}_{6}\left(\dfrac{3}{2}\right)^{9-6}\left(-\dfrac{1}{3}\right)^{6}+2 \cdot{ }^{9} \mathrm{C}_{7}\left(\dfrac{3}{2}\right)^{9-7}\left(-\dfrac{1}{3}\right)^{7}$

$=\dfrac{9 \times 8 \times 7 \times 6 !}{6 ! \times 3 \times 2} \cdot \dfrac{3^{3}}{2^{3}} \cdot \dfrac{1}{3^{6}}-2 \cdot \dfrac{9 \times 8 \times 7 !}{7 ! \times 2 \times 1} \cdot \dfrac{3^{2}}{2^{2}} \cdot \dfrac{1}{3^{7}}$

$=\dfrac{84}{8} \cdot \dfrac{1}{3^{3}}-\dfrac{36}{4} \cdot \dfrac{2}{3^{5}} .$

$=\dfrac{7}{18}-\dfrac{2}{27}$

$=\dfrac{21-4}{54}$

$=\dfrac{17}{54}$

Objective Type Question:

18. The total number of terms in the expansion of $(x+a)^{100}+(x-a)^{100}$ after simplification is

(A) $\mathbf{5 0}$

(B) 202

(C) 51

(D) None of these

Ans: Correct answer is option (C)

Given, $(x+a)^{100}-(x-a)^{100}$.

Number of terms in the expansion of $(x+a)^{n}+(x-a)^{n}$ is $(n+2)$.

The total number of the terms in the expansion of $(x+a)^{100}-(x-a)^{100}$ are 101 . Out of 101 terms 50 terms are cancelled out due to same magnitude and opposite sign. Rest 51 terms are added.

Therefore, total number of terms are $101-50=51$.

19. If the integer $r>1, n>2$ and coefficients of ${(3 r)}^{th}$ and ${(r+2)}^{th}$ is the binomial expansion of $(1+x)^{2 n}$ are equal, then

(A) $\mathrm{n}=2 \mathrm{r}$

(B) $\mathrm{n}=3 \mathrm{r}$

(C) $n=2 r+1$

(D) None of these

Ans: Correct answer is option (A)

Given, $(1+x)^{2 n}$

In expansion of $(x+y)^{n}$, the coefficient of $(r+2)^{th}$ term is ${ }^{n} C_{r}$.

For $(3r)^{th}$ term, $\mathrm{T}_{3 \mathrm{r}-1+1}$ is

$\mathrm{T}_{3 \mathrm{r}}={ }^{2 \mathrm{n}} \mathrm{C}_{3 \mathrm{r}-1} \mathrm{x}^{3 \mathrm{r}-1}$

For $(r+2)^{th}$ term, $T_{r+1+1}$ is

$\mathrm{T}_{\mathrm{r}+2}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+1} \mathrm{x}^{\mathrm{r}+1}$

$\text { As, }{ }^{2 \mathrm{n}} \mathrm{C}_{3 \mathrm{r}-1}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+1}$

$\Rightarrow 3 \mathrm{r}-1=\mathrm{r}+1$

$\Rightarrow 4 \mathrm{r}=2 \mathrm{n}$

$\Rightarrow \mathrm{n}=2 \mathrm{r}$

20. The two successive terms in the expansion of $(1+x)^{24}$ whose coefficients are in the ratio $1: 4$ are

(A) $3^{rd}$ and $4^{th}$

(B) $4^{th}$ and $5^{th}$

(C) $5^{th}$ and $6^{th}$

(D) $6^{th}$ and $7^{th}$

Ans: Correct answer is option (C)

Given, Ratio of successive terms $1: 4$.

Consider the two successive terms as $(r+1)^{th}$ and $(r+2)^{th}$. Then, use the formula of the general term.

$\mathrm{T}_{\mathrm{r}+1}={ }^{24} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}} \text { and } \mathrm{T}_{\mathrm{r}+2}={ }^{24} \mathrm{C}_{\mathrm{r}+1} \mathrm{x}^{\mathrm{r}+1}$

$\text { As, } \dfrac{{ }^{24} \mathrm{C}_{\mathrm{r}}}{{ }^{24} \mathrm{C}_{\mathrm{r}+1}}=\dfrac{1}{4}$

$\Rightarrow \dfrac{\dfrac{(24) !}{\mathrm{r} !(24-\mathrm{r}) !}}{\dfrac{(24) !}{(\mathrm{r}+1) !(24-\mathrm{r}-1) !}}=\dfrac{1}{4}$

$\Rightarrow \dfrac{(\mathrm{r}+1)(\mathrm{r}) !(23-\mathrm{r}) !}{\mathrm{r} !(24-\mathrm{r})(23-\mathrm{r}) !}=\dfrac{1}{4}$

$\Rightarrow \dfrac{\mathrm{r}+1}{24-\mathrm{r}}=\dfrac{1}{4}$

$\Rightarrow 4 \mathrm{r}+4=24-\mathrm{r}$

$\Rightarrow \mathrm{r}=4$

$\therefore(r+1)^{th}$ and $(r+2)^{th}$ terms are $5^{th}$ and $6^{th}$ terms.

21. The coefficient of $x^{n}$ in the expansion of $(1+x)^{2 n}$ and $(1+x)^{2 n-1}$ are in the ratio

(A) $1: 2$

(B) $1: 3$

(C) $3: 1$

(D) $2: 1$

Ans: Correct answer is option (D)

Given, $(1+x)^{2 n}$ and $(1+x)^{2 n-1}$.

First, find out the coefficient of $x^{\text {n. }}$. Then, divide them to obtain the ratio.

Coefficient of $x^{n}$ in $(1+x)^{2 n}={ }^{2 n} C_{n}$

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

$\therefore \dfrac{{ }^{2 n} C_{n}}{{ }^{2 n-1} C_{n}}=\dfrac{\dfrac{(2 n) !}{n ! n !}}{\dfrac{(2 n-1) !}{n !(n-1) !}}$

$=\dfrac{(2 n) ! n !(n-1) !}{n ! n !(2 n-1) !}$ $=\dfrac{(2 n)(2 n-1) !(n !)(n-1) !}{n !(n)(n-1) !(2 n-1) !}$

$=\dfrac{2 \mathrm{n}}{\mathrm{n}}$

$=\dfrac{2}{1}$

$=2: 1$

22. If the coefficient of $2^{nd}$, $3^{th}$, and $4^{th}$ term in expansion of $(1+x)^{n}$ are in AP, then the value of $n$ is

(A) 2

(B) 7

(C) 11

(D) 14

Ans: Correct answer is option (B)

Given, Coefficients of second, third and fourth are in AP.

In expansion of $(x+y)^{n}$, the coefficient of $(r+1)^{\text {th }}$ term is ${ }^{n} C_{r}$. If $a$, $b$ and $c$ are in AP, then $2 b=a+c$.

In $(1+x)^{n}$,

coefficient of $2^{nd}$ term $={ }^{n} C_{1}$

coefficient of $3^{rd}$ term $={ }^{n} C_{2}$

coefficient of fourth term $={ }^{n} C_{3}$

Since, coefficients are in AP.

$\therefore 2^{n} C_{2}={ }^{n} C_{1}+{ }^{n} C_{3}$

$\Rightarrow 2\left[\dfrac{n(n-1)(n-2) !}{2 \times 1 \times(n-2) !}\right]=\dfrac{n(n-1) !}{(n-1) !}+\dfrac{n(n-1)(n-2)(n-3) !}{3 !(n-3) !}$

$\Rightarrow n(n-1)=n+\dfrac{n(n-1)(n-2)}{6}$

$\Rightarrow 6 n-6=6+n^{2}-3 n+2$

$\Rightarrow n^{2}-7 n+2 n+14=0$

$\Rightarrow n(n-7)-2(n-7)=0$

$\Rightarrow(n-7)(n-2)=0$

$\therefore n=2(n o t$ possible $)$ or $n=7$

23. If $A$ and $B$ are coefficient of $x^{n}$ in expansion of $(1+x)^{2 n}$ and $(1+x)^{2 n-1}$ respectively, then $\dfrac{\mathrm{A}}{\mathrm{B}}$ equals to

(A) 1

(B) 2

(C) $\dfrac{1}{2}$

(D) $\dfrac{1}{n}$

Ans: Correct answer is option (B)

Given, $(1+x)^{2 n}$ and $(1+x)^{2 n-1}$.

First, find out the coefficient of $x^{n}$. Then, divide them to obtain the ratio.

Coefficient of $x^{n}$ in $(1+x)^{2 n}$ is $A={ }^{2 n} C_{n}$

Coefficient of $x^{n}$ in $(1+x)^{2 n-1}$ is $B={ }^{2 n-1} C_{n}$

$\therefore \dfrac{\mathrm{A}}{\mathrm{B}}=\dfrac{{ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}}}{{ }^{2 \mathrm{n}-1} \mathrm{C}_{\mathrm{n}}}=\dfrac{\dfrac{(2 \mathrm{n}) !}{\mathrm{n} ! \mathrm{n} !}}{\dfrac{(2 \mathrm{n}-1) !}{\mathrm{n} !(\mathrm{n}-1) !}}$

$\therefore \dfrac{\mathrm{A}}{\mathrm{B}}=\dfrac{{ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}}}{{ }^{2 \mathrm{n}-1} \mathrm{C}_{\mathrm{n}}}=\dfrac{\dfrac{(2 \mathrm{n}) !}{\mathrm{n} ! \mathrm{n} !}}{\dfrac{(2 \mathrm{n}-1) !}{\mathrm{n} !(\mathrm{n}-1) !}}$

$=\dfrac{(2 \mathrm{n}) ! \mathrm{n} !(\mathrm{n}-1) !}{\mathrm{n} ! \mathrm{n} !(2 \mathrm{n}-1) !}$

$=\dfrac{(2 \mathrm{n})(2 \mathrm{n}-1) !(\mathrm{n} !)(\mathrm{n}-1) !}{\mathrm{n} !(\mathrm{n})(\mathrm{n}-1) !(2 \mathrm{n}-1) !}$

$=\dfrac{2 \mathrm{n}}{\mathrm{n}}$

$=\dfrac{2}{1}$

$=2: 1$

24. If the middle term of $\left(\dfrac{1}{x}+x \sin x\right)^{10}$ is equal to $7 \dfrac{7}{8}$, then the value of $x$ is

(A) $2 \mathrm{n} \pi+\dfrac{\pi}{6}$

(B) $n \pi+\dfrac{\pi}{6}$

(C) $n \pi+(-1)^{n} \dfrac{\pi}{6}$

(D) $\mathrm{n} \pi+(-1)^{\mathrm{n}} \dfrac{\pi}{3}$

Ans: Correct answer is option (C)

Given, Middle term of expansion is $7 \dfrac{7}{8}$.

For even power of binomial, the middle term is $\left(\dfrac{\mathrm{n}}{2}+1\right)^{th}$  term. for $n=10$, middle term $=\left(\dfrac{10}{2}+1\right)^{th}$

$=6^{th}$  term

$\therefore \mathrm{T}_{6}=\mathrm{T}_{5+1}={ }^{10} \mathrm{C}_{5}\left(\dfrac{1}{\mathrm{x}}\right)^{10-5}(\mathrm{x} \sin \mathrm{x})^{5}$

$\Rightarrow \dfrac{63}{8}={ }^{10} \mathrm{C}_{5} \mathrm{x}^{-5} \mathrm{x}^{5} \sin ^{5} \mathrm{x}$

$\Rightarrow \dfrac{63}{8}=\dfrac{10.9 .8 .7 .6 .5 !}{5.4 .3 .2 .1 .5 !} \sin ^{5} \mathrm{x}$

$\Rightarrow \dfrac{63}{8}=2.9 .2 .7 . \sin ^{5} \mathrm{x}$

$\Rightarrow \sin ^{5} \mathrm{x}=\dfrac{1}{32}$

$\Rightarrow \sin ^{5} \mathrm{x}=\left(\dfrac{1}{2}\right)^{5}$

$\Rightarrow \sin \mathrm{x}=\dfrac{1}{2}$

$\therefore \mathrm{x}=\mathrm{n} \pi+(-1)^{\mathrm{n}} \dfrac{\pi}{6}$

Fill in the blanks:

25. The largest coefficient in the expansion of $(1+x)^{30}$ is_______

Ans: Given, $(1+x)^{30}$.

In expansion of $(1+x)^{n}$, the largest coefficient for even in $n$ is ${ }^{n} C_{\dfrac{n}{2}}$.

Therefore, the largest the coefficient is given by, $(1+x)^{30}={ }^{30} C_{\dfrac{30}{2}}={ }^{30} C_{15}$.

26. The number of terms in the expansion of $(x+y+z)^{n}$________

Ans: Given, $(x+y+z)^{n}$.

Expand $(x+y+z)^{n}=[x+(y+z)]^{n}$.

${[x+(y+z)]^{n}={ }^{n} C_{0} x^{n}+{ }^{n} C_{1} x^{n-1}(y+z)+\ldots \ldots \ldots+{ }^{n} C_{n}(y+z)^{n}}$

$\therefore \text { Number of terms }=1+2+3+4 \ldots \ldots .+n+(n-1)$

$=\dfrac{(n+1)(n+2)}{2}$

27. In the expansion of $\left(x^{2}-\dfrac{1}{x^{2}}\right)^{16}$, the value of constant term is______

Ans: Given, $\left(x^{2}-\dfrac{1}{x^{2}}\right)^{16}$.

Constant term is the term independent of $\mathrm{x}$. Use the concept of independent terms.

$\mathrm{T}_{\mathrm{r}+1}={ }^{16} \mathrm{C}_{\mathrm{r}}\left(\mathrm{x}^{2}\right)^{16-\mathrm{r}}\left(-\dfrac{1}{\mathrm{x}^{2}}\right)^{\mathrm{r}}$

$={ }^{16} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{32-2 \mathrm{r}}(-1)^{\mathrm{r}} \mathrm{x}^{-2 \mathrm{r}}$

$={ }^{16} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{32-4 \mathrm{r}}(-1)^{\mathrm{r}}$

for constant term, $32-4 \mathrm{r}=0$,

$\Rightarrow \mathrm{r}=8$

$\therefore \mathrm{T}_{8+1}={ }^{16} \mathrm{C}_{8}$

28. If the seventh term from beginning and the end in the expansion of $\left(\sqrt[3]{2}+\dfrac{1}{\sqrt[3]{3}}\right)^{n}$ are equal, then $n$ equals to_____.

Ans: Given, $\left(\sqrt[3]{2}+\dfrac{1}{\sqrt[3]{3}}\right)^{\mathrm{n}}$

Using the general term find the $7^{th}$ term from the beginning to the $7^{th}$ term from the end.

$7^{th}$ term from the beginning $\mathrm{T}_{7}=\mathrm{T}_{6+1}={ }^{\mathrm{n}} \mathrm{C}_{6}(\sqrt[3]{2})^{\mathrm{n}-6}\left(\dfrac{1}{\sqrt[3]{3}}\right)^{6}$

$7^{th}$ term from the end $=\mathrm{T}_{7}$ from beginning of $\left(\dfrac{1}{\sqrt[3]{3}}+\sqrt[3]{2}\right)^{\mathrm{n}}$

$\therefore \mathrm{T}_{7}={ }^{\mathrm{n}} \mathrm{C}_{6}\left(\dfrac{1}{\sqrt[3]{3}}\right)^{\mathrm{n}-6}(\sqrt[3]{2})^{6}$

As, ${ }^{\mathrm{n}}$ $\mathrm{C}_{6}(\sqrt[3]{2})^{\mathrm{n}-6}\left(\dfrac{1}{\sqrt[3]{3}}\right)^{6}={ }^{\mathrm{n}}$ $\mathrm{C}_{6}\left(\dfrac{1}{\sqrt[3]{3}}\right)^{\mathrm{n}-6}(\sqrt[3]{2})^{6}$

$\Rightarrow 2^{\dfrac{n-6}{3}} \cdot 3^{-\dfrac{6}{3}}=3^{-\left(\dfrac{n-6}{3}\right)} \cdot 2^{\dfrac{6}{3}}$

$\Rightarrow 2^{\dfrac{n-12}{3}}=\left(\dfrac{1}{3}\right)^{\dfrac{n-12}{3}}$

This is possible, when $\dfrac{\mathrm{n}-12}{3}=0$

$\Rightarrow \mathrm{n}=12$

29. The coefficient of $a^{-6} b^{4}$ in the expansion of $\left(\dfrac{1}{a}-\dfrac{2 b}{3}\right)^{10}$ is_______

Ans: Given, $\left(\dfrac{1}{a}-\dfrac{2 b}{3}\right)^{10}$

Use the general term $T_{r+1}={ }^{n} C_{r} x^{n-r}(-a)^{r}$ in expansion of $(x-a)^{n}$.

$\mathrm{T}_{\mathrm{r}+1}={ }^{10}$ $\mathrm{C}_{\mathrm{r}}\left(\dfrac{1}{\mathrm{a}}\right)^{10-\mathrm{r}}\left(-\dfrac{2 \mathrm{~b}}{3}\right)^{\mathrm{r}}$

Coefficient of $a^{-6} b^{4}, 10-r=6$

$\Rightarrow \mathrm{r}=4$

$\therefore$ Coefficient of $\mathrm{a}^{-6} \mathrm{~b}^{4}={ }^{10}$ $\mathrm{C}_{4}\left(-\dfrac{2}{3}\right)^{4}$

$=\dfrac{10.9 .8}{6 ! 4.3}$

$=\dfrac{1120}{27}$

30. Middle term in the expansion of $\left(a^{3}+b a\right)^{28}$ is______

Ans: Given, $\left(a^{3}+b a\right)^{28}$

Middle term for even power of binomial, $\left(\dfrac{\mathrm{n}}{2}+1\right)^{th}$ term.

Middle term $=\left(\dfrac{28}{2}+1\right)^{th}$ $=15^{th}$ term

$\therefore \mathrm{T}_{15}=\mathrm{T}_{14+1}$

$={ }^{28} \mathrm{C}_{14}\left(\mathrm{a}^{3}\right)^{28-14}(\mathrm{ba})^{14}$

$={ }^{28} \mathrm{C}_{14} \mathrm{a}^{42} \mathrm{~b}^{14} \mathrm{a}^{14}$

$={ }^{28} \mathrm{C}_{14} \mathrm{a}^{56} \mathrm{~b}^{14}$

31. The ratio of coefficient of $\mathbf{x}^{p}$ and $\mathbf{x}^{4}$ in the expansion of $r$ is_______

Ans: Given, $(1+\mathrm{x})^{\mathrm{p}+\mathrm{q}}$.

Coefficient of $r^{\text {th }}$ term in expansion of $(1+x)^{p+q}$ is ${ }^{p+q} C_{r}$.

Coefficient of $x^{p}$ in $(1+x)^{p+q}={ }^{p+q} C_{p}$

Coefficient of $x^{q}$ in $(1+x)^{p+q}={ }^{p+q} C_{q}$

$\therefore \dfrac{{ }^{p+q} C_{p}}{p^{p+q} C_{q}}=\dfrac{{ }^{p+q} C_{p}^{p+q} C_{p}}{p}=1: 1$

32. The position of the term independent of $x$ in expansion of $\left(\sqrt{\dfrac{x}{3}}+\dfrac{3}{2 x^{2}}\right)^{10}$________

Ans: Given, $\left(\sqrt{\dfrac{x}{3}}+\dfrac{3}{2 x^{2}}\right)^{10}$

For the term independent of $\mathrm{x}$ substitute the power of $\mathrm{x}$ equal to zero. Use general term.

$\mathrm{T}_{\mathrm{r}+1}={ }^{10} \mathrm{C}_{\mathrm{r}}\left(\sqrt{\dfrac{\mathrm{x}}{3}}\right)^{10-\mathrm{r}}\left(\dfrac{3}{2 \mathrm{x}^{2}}\right)^{\mathrm{r}}$

$={ }^{10} \mathrm{C}_{\mathrm{r}} \cdot \mathrm{x}^{\dfrac{10-\mathrm{r}}{2}} \cdot 3^{\dfrac{-10+\mathrm{r}}{2}} \cdot 3^{\mathrm{r}} \cdot 2^{-\mathrm{r}} \cdot \mathrm{x}^{-2 \mathrm{r}}$

$={ }^{10} \mathrm{C}_{\mathrm{r}} \cdot \mathrm{x}^{\dfrac{10-5 \mathrm{r}}{2}} \cdot 3^{\dfrac{-10+3 \mathrm{r}}{2}} \cdot 2^{-\mathrm{r}}$

For term independent of $x, \dfrac{10-5 r}{2}=0$

$\Rightarrow \mathrm{r}=2$

$\therefore \mathrm{T}_{\mathrm{r}+1}=\mathrm{T}_{2+1}=\mathrm{T}_{3}$

33. If $25^{15}$ is divisible by 13 , then the remainder is______

Ans: Given, $25^{15}$ is divisible by 13 .

Expand $25^{15}=(26-1)^{15}$. Separate the terms divisible by 13 . The rest of the term will be the remainder.

$25^{15}=(26-1)^{15}$

$={ }^{15} \mathrm{C}_{0} 26^{15}-{ }^{15} \mathrm{C}_{1} 26^{14}+\ldots \ldots .-{ }^{15} \mathrm{C}_{15}$

$={ }^{15} \mathrm{C}_{0} 26^{15}-{ }^{15} \mathrm{C}_{1} 26^{14}+\ldots \ldots-1-13+13$

$={ }^{15} \mathrm{C}_{0} 26^{15}-{ }^{15} \mathrm{C}_{1} 26^{14}+\ldots .-13+12$

$=13 \mathrm{k}+12\left(\mathrm{k}={ }^{15} \mathrm{C}_{0} 26^{15}-{ }^{15} \mathrm{C}_{1} 26^{14}+\ldots .-13\right)$

$\therefore \text { Remainder }=12$

True or False:

34. The sum of the series $\sum_{r=0}^{10}{ }^{20} C_{r}$ is $2^{19}+\dfrac{{ }^{20} C_{10}}{2}$.

Ans: False.

Given, The sum of the series $\sum_{\mathrm{r}=0}^{10}{ }^{20} \mathrm{C}_{\mathrm{r}}$ is $2^{19}+\dfrac{{ }^{20} \mathrm{C}_{10}}{2}$.

Expand the given summation series as binomial expansion.

$\sum_{\mathrm{r}=0}^{10}{ }^{20} \mathrm{C}_{\mathrm{r}}={ }^{20} \mathrm{C}_{0}+{ }^{20} \mathrm{C}_{1}+{ }^{20} \mathrm{C}_{2}+\ldots \ldots+{ }^{20} \mathrm{C}_{10}$

$={ }^{20} \mathrm{C}_{0}+{ }^{20} \mathrm{C}_{1}+\ldots . .+{ }^{20} \mathrm{C}_{10}+{ }^{20} \mathrm{C}_{11}+\ldots . .+{ }^{20} \mathrm{C}_{20}-\left({ }^{20} \mathrm{C}_{11}+\ldots \ldots+{ }^{20} \mathrm{C}_{20}\right)$

$=2^{20}-\left({ }^{20} \mathrm{C}_{11}+\ldots \ldots+{ }^{20} \mathrm{C}_{20}\right)$

35. The expression $7^{9}+9^{7}$ is divisible by 64

Ans: True.

Given, The expression $7^{9}+9^{7}$ is divisible by 64 .

Use $(1+8)^{7}+(1-8)^{9}$. Separate the terms divisible by 64 . The rest term will be remainder.

$7^{9}+9^{7}=(1+8)^{7}+(1-8)^{9}$

$=\left({ }^{7} \mathrm{C}_{0}+{ }^{7} \mathrm{C}_{1} 8+{ }^{7} \mathrm{C}_{2} 8^{2}+\ldots \ldots .+{ }^{7} \mathrm{C}_{7} 8^{7}\right)-\left({ }^{9} \mathrm{C}_{0}-{ }^{9} \mathrm{C}_{1} 8+{ }^{9} \mathrm{C}_{2} 8^{2}+\ldots \ldots-{ }^{9} \mathrm{C}_{9} 8^{9}\right)$

$=\left(1+7 \times 8+21 \times 8^{2}+\ldots . .\right)-\left(1-9 \times 8+36 \times 8^{2}-\ldots \ldots . .8^{9}\right)$

$=(7 \times 8+9 \times 8)+\left(21 \times 8^{2}-36 \times 8^{2}\right)$

$=2 \times 64+(21-36) 64+\ldots \ldots \ldots .$

The expression is divisible by 64 .

36. The number of terms in the expansion of $\left[\left(2 x+y^{3}\right)^{4}\right]^{7}$ is 8 .

Ans: False.

Given, The number of terms in the expansion of $\left[\left(2 x+y^{3}\right)^{4}\right]^{7}$ is 8 .

The binomial expansion $(1+x)^{n}$ has $(n+1)$ terms. First, simplify the expression.

$\left[\left(2 x+y^{3}\right)^{4}\right]^{7}=\left(2 x+y^{3}\right)^{28}$

$\therefore$ The number of terms are 29 .

37. The sum of coefficients of two middle terms in the expansion of $(1+x)^{2 n-1}$ is equal to ${ }^{2 n-1} C_{n}$.

Ans: False.

Given, The sum of coefficients of two middle terms in the expansion of $(1+x)^{2 n-1}$ is equal to ${ }^{2 n-1} C_{n}$.

For the odd power of binomial, there exists two middle terms.

$\left(\dfrac{\mathrm{n}+1}{2}\right)^{th}$ and $\left(\dfrac{\mathrm{n}+1}{2}+1\right)^{th}$ term.

For $(1+x)^{2 n-1}$,

Middle terms are $\left(\dfrac{2 n-1+1}{2}\right)=n^{th}$ and $\left(\dfrac{2 n-1+1}{2}+1\right)=(n+1)^{th}$ term.

coefficient of $n^{th}$ term $={ }^{2 n-1} C_{n-1}$

coefficient of $(\mathrm{n}+1)^{th}$ term $={ }^{2 \mathrm{n}-1} \mathrm{C}_{\mathrm{n}}$

$\therefore$ Sum of coefficients $={ }^{2 n-1} C_{n-1}+{ }^{2 n-1} C_{n}$

$={ }^{2 n} C_{n} \quad\left(\because{ }^{n} C_{r}+{ }^{n} C_{r-1}={ }^{n+1} C_{r}\right)$

38. The last two digits of the numbers $3^{400}$ are 01 .

Ans: True.

Given, The last two digits of the numbers $3^{400}$ are 01 .

Expand using $3^{400}=9^{200}=(10-1)^{200}$.

$3^{400}=9^{200}=(10-1)^{200}$

$\Rightarrow(10-1)^{200}={ }^{200} \mathrm{C}_{0} 10^{200}-{ }^{200} \mathrm{C}_{1} 10^{199}+\ldots .-{ }^{200} \mathrm{C}_{199} 10^{1}+{ }^{200} \mathrm{C}_{200} 1^{200}$

$\Rightarrow(10-1)^{200}=10^{200}-200 \times 10^{199}+\ldots . .-10 \times 200+1$

Clearly, the last two digits are 01 .

39. If the expansion of $\left(x-\dfrac{1}{x^{2}}\right)^{2 n}$ contains a term independent of $x$, then $n$ is a multiple of 2 .

Ans: False.

Given, The expansion of $\left(x-\dfrac{1}{x^{2}}\right)^{2 n}$ contains a term independent of $\mathrm{x}$, then $\mathrm{n}$ is a multiple of 2 .

Use the general term in expansion of $(x-a)^{n}$ is $T_{r+1}={ }^{n} C_{r}(x)^{n-r}(-a)^{r}$.

The term independent of $x$ is obtained by substituting $n-r=0$.

$T_{r+1}={ }^{2 n} C_{r}(x)^{2 n-r}\left(-\dfrac{1}{x^{2}}\right)^{r}$

$={ }^{2 n} C_{r}(x)^{2 n-r}(-1)^{r} x^{-2 r}$

$={ }^{2 n} C_{r}(x)^{2 n-3 r}(-1)^{r}$

For independent term, $2 n-3 r=0$

$\Rightarrow \mathrm{r}=\dfrac{2 \mathrm{n}}{3}($ not possible $)$

40. The number of terms in the expansion of $(a+b)^{n}$, $n \in N$, is one less than the power of $\mathrm{n}$.

Ans: False.

Given, Number of terms in the expansion of $(a+b)^{n}$, where $n \in N$, is one less than the power of $n$.

In the binomial expansion, $(x+y)^{n}$ there are $(n+1)$ terms.

The number of terms in the expansion of $(a+b)^{n}$, where $n \in N$, is $(n+1)$. So, the number of terms are one more than power of $\mathrm{n}$.

An Introduction to Binomial Theorem - Class 11 Maths Chapter 8

One of the most important and scoring topics in Chapter 8 of NCERT Class 11 Maths- Binomial Theorem. The main purpose of using the Binomial Theorem is to find or calculate the expanded value of the algebraic expression (x+y)n where the sum of the terms involving individual exponents of variables x and y are deduced. The coefficient is the numeric value associated with each term in the binomial expansion.

Binomial theorem works on expressions in which the exponential value is 4 and above negative and non-negative values or even fractions.  As stated in the binomial theorem, the expression (x+y)n can be expanded into a sum in the form of axbyc where b and c are non-negative integers and b+c=n  and the coefficient an in each term of the expression is non-negative/ specific positive integer depending on the values of n and b. If the exponent is zero, then the value of the variable with corresponding power gets omitted.

For example, let us consider the following expression:

(x+y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4

Another way of finding the sum of coefficients in binomial expansion is Pascal’s triangle in which a diagram helps to determine the binomial expansion here instead of algebraic expansion. Here each number in the triangle is the sum of the two directly above it. Each row of Pascal’s Triangle is numbered, starting from row 1 as n=0. Then keep adding the two above numbers to find a new value and in case of no values present, place a zero.The Rows Continue Indefinitely

NCERT Exemplar for Class 11 Maths Chapter 8- Binomial theorem is divided into sections from 8.1 to 8.1.7. Each section defines and explains topics as follows.

• The Binomial theorem definition;

• Certain observations (like, the total number of terms is n+1 or 1 more than the exponent n; the sum of indices of a and b equals n, i.e., the power of binomial; Use of Pascal’s Triangle)

• Some particular cases like using positive integers;

• The path term at the end in the expression of (x+y)n is (n-p+2)th from the beginning;

• Middle Terms in case of n being odd or even;

• Binomial Coefficients.

In order to understand the different concepts of the binomial theorem and to utilize them to solve problems, download this solution file for NCERT Exemplar Class 11 Maths Chapter 8 from Vedantu. Find out how the experts have designed the solutions following the CBSE guidelines and made it easier for you. Practice following these solutions and save your precious time. Learn how to approach such problems and focus on scoring more marks with this solution file.

FAQs on NCERT Exemplar for Class 11 Maths Chapter 8 - Binomial Theorem (Book Solutions)

1. Is NCERT Exemplar for Class 11 Maths Chapter 8 - Binomial Theorem (Book Solutions) easy to study?

The chapter Binomial Theorem is very easy due to its simpler concepts. It can be extremely scoring if one studies and understands the chapter very carefully. Because the expressions and formulae can be confusing initially, practicing regularly while stepping up the level one by one can be helpful for the students. The NCERT Exemplar for Class 11 is an excellent book for clearing concepts in the standard form. Students can practice examples as well as a set of question banks given after the Chapter.

2. How many exercises are given in NCERT Exemplar for Class 11 Maths Chapter 8- Binomial Theorem?

NCERT is known for its concept-based lessons which are easy to learn and understand. NCERT Class 11 Maths has question banks after every chapter of decent difficulty which students can practice regularly and revise the lessons accordingly. NCERT Exemplar for Class 11 Maths Chapter 8- Binomial Theorem has one exercise (8.3) which comprises questions divided into short answer types, long answer types, and Multiple Choice questions. Students can check out some more question banks/sample papers by Vedantu for free.

3. Should students practise examples of Binomial Theorem first?

To put the learning to the test, students must go through the examples first before moving on to the question banks. Students often skip the examples and directly move on to solving the problems in the exercise at the end of the chapters. This leads to confusion since students are unaware of how to solve and which methods to use to find the solutions to the questions. Hence practicing the examples first is helpful for complete clarity of the Chapter. NCERT Exemplar for Class 11 Maths Chapter 8- Binomial Theorem (Book Solutions) can be found on the Vedantu website as well.

4. Is Binomial Theorem important for JEE?

The binomial theorem is difficult for only those students who have been unable to clear doubts or understand the chapter in the first place. There are different examples of varied difficulty levels given in between the Chapters to help the student put their knowledge into use and to revise accordingly. For JEE, the Binomial theorem exists as an important aspect of Algebra which is a scoring portion in the paper. Students can easily score full marks by understanding, learning, and practicing different formulae every day.

5. How do we learn the formulae of the Binomial Theorem?

Formulae in Maths have to be practiced and revised regularly to reduce pressure during the exam. The general formula of the Binomial Theorem can be confusing to understand and learn because it is lengthy with several terms. So, the best way to familiarise oneself with the formula is through derivation. Short notes can be made to remember it for a longer period. Use different methods of study and online platforms like Vedantu, video tutorials, etc to clear doubts.