# NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.2

## Chapter 8 Ex.8.2 Question 1

Find the coefficient of $${x^5}$$ in$${\left(x + 3\right)^8}.$$

### Solution

It is known that $${\left(r + 1\right)^{th}}$$ term,$$\left( {T_r + 1} \right)$$ in the binomial expression of $${\left(a + b\right)^n}$$ is given by

${T_r + 1} = {}^n{C_r}{a^{n - r}}{b^r}$

Assuming $${x^5}$$ occurs in the expansion of $${\left(x + 3\right)^8},$$ we obtain

${T_r + 1} = {}^8{C_r}{\left( x \right)^{8 - r}}{\left( 3 \right)^r}$

Comparing the indices of $$x$$ in $$x^5$$ in $$\left( {T_r + 1} \right),$$ we obtain

\begin{align}8 - r &= 5\\r &= 3\end{align}

Thus, the coefficient of $$x^5$$is $${}^8{C_3}{\left( 3 \right)^3}$$

\begin{align}{}^8{C_3}{\left( 3 \right)^3} &= \frac{{8!}}{{3!\left( 5 \right)!}} \times {\left( 3 \right)^3}\\ &= \frac{{8 \times 7 \times 6 \times \left( {5!} \right)}}{{3 \times \left( {2!} \right) \times \left( {5!} \right)}} \times 27\\ &= 1512\end{align}

## Chapter 8 Ex.8.2 Question 2

Find the coefficient of $$a^5b^7$$ in $${\left( a - 2b \right)^{12}}$$

### Solution

It is known that $${\left(r + 1\right)^{th}}$$ term, $$\left( T_r + 1\right)$$ in the binomial expression of $${\left(a + b\right)^n}$$ is given by $${T_r + 1} = {}^n{C_r}{a^n - r}{b^r}$$

Assuming $$a^5b^7$$ occurs in the expansion of $$\left( a - 2b \right)^{12}$$, we obtain

\begin{align}{T_r + 1} &= {}^{12}{C_r}{\left( a \right)^{12 - r}}{\left( { - 2b} \right)^r}\\ &= {}^{12}{C_r}{\left( { - 2} \right)^r}{\left( a \right)^{12 - r}}{\left( b \right)^r}\end{align}

Comparing the indices of $$a$$ and $$b$$ in $${a^5b^7}$$ in $$\left( {T_r + 1} \right),$$ we obtain

$r = 7$

Thus, the coefficient of $$a^5b^7$$ is $${}^{12}{C_7}{\left( { - 2} \right)^7}$$

\begin{align}{}^{12}{C_7}{\left( { - 2} \right)^7} &= \frac{{12!}}{{7!\left( 5 \right)!}} \times {\left( { - 2} \right)^7}\\ &= \frac{{12 \times 11 \times 10 \times 9 \times 8 \times \left( {7!} \right)}}{{\left( {7!} \right) \times \left( {5!} \right)}} \times \left( { - 128} \right)\\ &= - \left( {792} \right) \times \left( {128} \right)\\& = - 101376\end{align}

## Chapter 8 Ex.8.2 Question 3

Write the general term in the expansion of $${\left(x^2 - y\right)^6}$$

### Solution

It is known that $${\left(r + 1\right)^{th}}$$ term, $$\left(T_r + 1\right)$$ in the binomial expression of $${\left(a + b\right)^n}$$ is given by $${T_r + 1} = {}^n{C_r}{a^{n - r}}{b^r}$$

Thus, the general term in the expansion of $$\left( x^2 - y \right)^6$$ is

\begin{align}{T_{r + 1}} &= {}^6{C_r}{\left( x^2 \right)^{6 - 4}}{\left( - y \right)^r}\\ &= {\left( - 1 \right)^r}{}^6{C_r}{\left( x \right)^{12 - 2r}}{\left( y \right)^r}\end{align}

## Chapter 8 Ex.8.2 Question 4

Write the general term in expansion of $${\left( x^2 - yx \right)^{12}},\;x \ne 0$$

### Solution

It is known that $${\left(r + 1 \right)^{th}}$$ term, $$\left(T_{r + 1} \right)$$ in the binomial expression of $${\left(a + b\right)^n}$$ is given by $$T_{r + 1} = {}^n{C_r}{a^{n - r}}{b^r}$$

Thus, the general term in the expansion of $${\left( {{x^2} - yx} \right)^{12}}$$ is

\begin{align}{T_{r + 1}} &= {}^{12}{C_r}{\left( {{x^2}} \right)^{12 - r}}{\left( { - yx} \right)^r}\\ &= {}^{12}{C_r}{\left( x \right)^{24 - 2r}}{\left( { - 1} \right)^r}{\left( y \right)^r}{\left( x \right)^r}\\ &= {\left( { - 1} \right)^r}{}^{12}{C_r}{\left( x \right)^{24 - r}}{\left( y \right)^r}\end{align}

## Chapter 8 Ex.8.2 Question 5

Find the $$4^{th}$$ term in the expansion of $${\left(x - 2y\right)^{12}}$$

### Solution

It is known that $${\left( {r + 1} \right)^{th}}$$term, $$\left( {T_{r + 1}} \right)$$ in the binomial expression of $${\left(a + b\right)^n}$$ is given by $${T_r + 1} = {}^n{C_r}{a^{n - r}}{b^r}$$

Thus, the $$4^{th}$$ term in the expansion of $${\left(x - 2y\right)^{12}}$$ is

\begin{align}{T_4} &= {T_{3 + 1}}\\ &= {}^{12}{C_3}{\left( x \right)^{12 - 3}}{\left( { - 2y} \right)^3}\\& = \frac{{12!}}{{3!9!}}{\left( x \right)^9}{\left( { - 2} \right)^3}{\left( y \right)^3}\\ &= \frac{{12 \times 11 \times 10}}{{3 \times 2}} \times \left( { - 8} \right){x^9}{y^3}\\ &= - 1760{x^9}{y^3}\end{align}

## Chapter 8 Ex.8.2 Question 6

Find the $$13^{th}$$ term in the expansion of $${\left( 9x - \frac{1}{{3\sqrt x }} \right)^{18}},\;x \ne 0$$

### Solution

It is known that $${\left(r + 1\right)^{th}}$$ term, $$(T_{r + 1})$$ the binomial expression of $$( {a + b})^n$$ is given by $$T_{r + 1} = {}^n{C_r}{a^{n - r}}{b^r}$$

Thus, the $$13^{th}$$ term in the expansion of $${\left( {9x - \frac{1}{{3\sqrt x }}} \right)^{18}},\;x \ne 0$$

\begin{align} {{T}_{13}}&={{T}_{12+1}} \\ & ={}^{18}{{C}_{12}}{{\left( 9x \right)}^{18-12}}{{\left( -\frac{1}{3\sqrt{x}} \right)}^{12}} \\ & =\frac{18!}{\left( 12! \right)\left( 6! \right)}\times {{9}^{6}}\times {{\left( x \right)}^{6}}{{\left( -\frac{1}{3} \right)}^{12}}{{\left( \frac{1}{\sqrt{x}} \right)}^{12}} \\ & =\frac{18\times 17\times 16\times 15\times 14\times 13\times \left( 12! \right)}{\left( 12! \right)\times 6\times 5\times 4\times 3\times 2}\times \left( {{3}^{12}} \right)\times \left( \frac{1}{{{3}^{12}}} \right)\times \left( {{x}^{6}} \right)\times \left( \frac{1}{{{x}^{6}}} \right)\\&\qquad \qquad \left[ \because {{9}^{6}}={{\left( {{3}^{2}} \right)}^{6}}={{3}^{12}} \right] \\ & =18564\end{align}

## Chapter 8 Ex.8.2 Question 7

Find the middle terms in the expansion of $${\left(3 - \frac{x^3} It is known that in the expansion of \(({a + b})^n,$$ when $$n$$ is odd; there are two middle terms, namely $$\left({\frac{{n + 1}}{2}}\right)^{th}$$ term and $$\left({\frac{{n + 1}}{2} + 1}\right)^{th}$$ term.

Therefore, the middle terms in the expansion $$\left({3 - \frac{x^3}{6}}\right)^7$$ are $${\left( {\frac{{7 + 1}}{2}} \right)^{th}} = {4^{th}}$$ term and $${\left( {\frac{{7 + 1}}{2} + 1} \right)^{th}} = {5^{th}}$$ term

\begin{align}{T_4} &= {T_{3 + 1}}\\ &= {}^7{C_3}{\left( 3 \right)^{7 - 3}}{\left( { - \frac{x^3}{6}} \right)^3}\\ &= \frac{{\left( {7!} \right)}}{{\left( {3!} \right)\left( {4!} \right)}} \times \left( {{3^4}} \right) \times {\left( { - 1} \right)^3} \times \left( {\frac{x^9}{6^3}} \right)\\ &= - \frac{{7 \times 6 \times 5 \times \left( {4!} \right)}}{{3 \times 2 \times \left( {4!} \right)}} \times \left( {{3^4}} \right) \times \left( {\frac{1}{{2^3 \times 3^3}}} \right) \times \left( {x^9} \right)\\ &= - \frac{105}{8}{x^9}\end{align}

Now,

\begin{align}{T_5}& = {T_{4 + 1}}\\& = {}^7{C_4}{\left( 3 \right)^{7 - 4}}{\left( { - \frac{x^3}{6}} \right)^4}\\ &= \frac{{7!}}{{\left( {4!} \right)\left( 3 \right)!}} \times {\left( 3 \right)^3} \times \left( {\frac{x^{12}}{6^4}} \right)\\ &= \frac{{7 \cdot 6 \cdot 5 \cdot 4!}}{{3 \cdot 2!\left( 4 \right)!}} \times \left( 3^3\right) \times \left( {\frac{1}{{2^4} \times {3^4}}} \right) \times \left(x^{12} \right)\\ &= \frac{35}{48}{x^{12}}\end{align}

Thus, the middle terms in the expansion of $${\left( {3 - \frac{x^3}{6}} \right)^7}$$ are \begin{align} - \frac{105}{8}{x^9}\end{align} and \begin{align}\frac{35}{48}{x^{12}}\end{align}

\right)^7}\)

{6}

## Chapter 8 Ex.8.2 Question 8

Find the middle terms in the expansion of $$\left( {\frac{x}{3} + 9y} \right)^{10}$$

### Solution

It is known that in the expansion of $$\left( {a + b} \right)^n,$$ when $$n$$ is even; the middle term is $$\left( {\frac{n}{2} + 1} \right)^{th}$$ term.

Therefore, the middle term in the expansion of $$\left( {\frac{x}{3} + 9y} \right)^{10}$$ is $${\left( {\frac{10}{2} + 1} \right)^{th}} = 6^{th}$$ term

\begin{align}{T_6} &= {T_{5 + 1}}\\ &= {}^{10}{C_5}{\left( {\frac{x}{3}} \right)^{10 - 5}}{\left( {9y} \right)^5}\\& = \frac{{\left( {10!} \right)}}{{\left( {5!} \right)\left( {5!} \right)}} \times \left( {\frac{{x^5}}{3^5}} \right) \times \left( {{9^5}} \right) \times \left( {y^5} \right)\\ &= \frac{{10 \times 9 \times 8 \times 7 \times 6 \times \left( {5!} \right)}}{{5 \times 4 \times 3 \times 2 \times \left( {5!} \right)}} \times \left( {\frac{1}{{3^5}}} \right) \times \left( {3^{10}} \right) \times \left( {x^5} \right)\left( {y^5} \right)\\ &= 252 \times {3^5} \times {x^5} \times {y^5}\\ &= 61236{x^5}{y^5}\end{align}

## Chapter 8 Ex.8.2 Question 9

In the expansion of $$\left( {1 + a} \right)^{m + n},$$ prove that the coefficients of $$a^m$$ and $$a^n$$ are equal.

### Solution

It is known that $$\left( {r + 1} \right)^{th}$$ term, $$\left( T_{r + 1} \right)$$ the binomial expression of $$\left(a + b\right)^n$$ is given by

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

Assuming that $$a^m$$ occurs in the $$\left(r + 1\right)^{th}$$ term of the expansion $$\left( 1 + a \right)^{m + n},$$ we obtain

${T_{r + 1}} = {}^{m + n}{C_r}{\left( 1 \right)^{m + n - r}}{\left( a \right)^r} = {}^{m + n}{C_r}{a^r}$

Comparing the indices of $$a$$ in $$a^m$$and in $$T_{r + 1},$$ we obtain $$r = m$$

Therefore, the coefficient of $$a^m$$ is

\begin{align}{}^{m + n}{C_m} &= \frac{{\left( {m + n} \right)!}}{{m!\left( {m + n - m} \right)!}}\\ &= \frac{{\left( {m + n} \right)!}}{{\left( {m!} \right)\left( {n!} \right)}}\qquad \qquad \ldots \left( 1 \right)\end{align}

Assuming that $$a^n$$ occurs in the $$\left( {k + 1} \right)^{th}$$ term of the expansion $$\left( {1 + a} \right)^{m + n},$$ we obtain

${T_{k + 1}} = {}^{m + n}{C_k}{\left( 1 \right)^{m + n - k}}{\left( a \right)^k} = {}^{m + n}{C_k}{a^k}$

Comparing the indices of $$a$$ in $$a^n$$ and in $$T_{k + 1},$$ we obtain $$k = n$$

Therefore, the coefficient of $${a^n}$$ is

\begin{align}{}^{m + n}{C_n} &= \frac{{\left( {m + n} \right)!}}{{n!\left( {m + n - n} \right)!}}\\ &= \frac{{\left( {m + n} \right)!}}{{\left( {m!} \right)\left( {n!} \right)}} \qquad \quad \ldots \left( 2 \right)\end{align}

Thus from (1) and (2), it is clear that the coefficients of $$a^m$$ and $$a^n$$ in the expansion of $$\left( {1 + a} \right)^{m + n}$$ are equal

Hence proved.

## Chapter 8 Ex.8.2 Question 10

The coefficients of the $$\left( r - 1 \right)^{th},\;r^{th}$$ and $$\left( {r + 1} \right)^{th}$$ term in the expansion of $$\left( {x + 1} \right)^n$$ are in the ratio of $$1:3:5$$. Find $$n$$ and $$r$$.

### Solution

It is known that $$\left( {k + 1} \right)^{th}$$ term, ($$T_{k + 1}$$) term of the expansion $$\left( {a + b} \right)^n$$ is given by

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

Hence, $$\left( {r - 1} \right)^{th}$$ term in the expansion of $$\left( {x + 1} \right)^n$$ is

\begin{align}T_{r - 1} &= {}^n{C_{r - 2}}\left( x \right)^{n - \left( {r - 2} \right)}\left( 1 \right)^{\left( {r - 2} \right)}\\ &= {}^n{C_{r - 2}}{x^{n - r + 2}}\end{align}

$$\left( {r + 1} \right)^{th}$$ term in the expansion of  $$\left( {x + 1} \right)^n$$ is

\begin{align}{T_{r + 1}} &= {}^n{C_r}{\left( x \right)^{n - r}}{\left( 1 \right)^r}\\ &= {}^n{C_r}{x^{n - r}}\end{align}

$$r^{th}$$ term in the expansion of $$\left( {x + 1} \right)^n$$ is

\begin{align}{T_r} &= {}^n{C_{r - 1}}{\left( x \right)^{n - \left( {r - 1} \right)}}{\left( 1 \right)^{\left( {r - 1} \right)}}\\& = {}^n{C_{r - 1}}{x^{n - r + 1}}\end{align}

Therefore, coefficients of the $$\left( {r - 1} \right)^{th},\;r^{th}$$ and $$\left( {r + 1} \right)^{th}$$ term in the expansion of $$\left( {x + 1} \right)^n$$ are $${}^n{C_{r - 2}}$$, $${}^n{C_{r - 1}}$$ and $${}^n{C_r}$$ respectively.

Since these coefficients are in the ratio of $$1:3:5$$, we obtain

$$\frac{{}^n{C_{r - 2}}}{{{}^n{C_{r - 1}}}} = \frac{1}{3}$$ and $$\frac{{}^n{C_{r - 1}}}{{}^n{C_r}} = \frac{3}{5}$$

\begin{align}\frac{{}^n{C_{r - 2}}}{{}^n{C_{r - 1}}} &= \frac{{n!}}{{\left( {r - 2} \right)!\left( {n - r + 2} \right)!}} \times \frac{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}}{{n!}}\\\frac{1}{3} &= \frac{{\left( {r - 1} \right).\left( {r - 2} \right)!\left( {n - r + 1} \right)!}}{{\left( {r - 2} \right)!\left( {n - r + 2} \right)!\left( {n - r + 1} \right)!}}\\\frac{1}{3} &= \frac{{r - 1}}{{n - r + 2}}\\n - r + 2 &= 3r - 3\\n - 4r + 5 &= 0\\n &= 4r - 5 \qquad \ldots \left( 1 \right)\end{align}

\begin{align}\frac{{}^n{C_{r - 1}}}{{}^n{C_r}} &= \frac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}} \times \frac{{r!\left( {n - r} \right)!}}{{n!}}\\\frac{3}{5} &= \frac{{r.\left( {r - 1} \right)!\left( {n - r} \right)!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!\left( {n - r} \right)!}}\\\frac{3}{5} &= \frac{r}{{n - r + 1}}\\3n - 3r + 3& = 5r\\3n - 8r + 3&= 0 \qquad \qquad \ldots \left( 2 \right)\end{align}

From (1) and (2), we obtain

\begin{align}3\left( {4r - 5} \right) - 8r + 3 &= 0\\12r - 15 - 8r + 3 &= 0\\4r - 12 &= 0\\r &= 3\end{align}

Putting the value of $$r$$ in (1), we obtain $$n$$

\begin{align}n &= 4 \times 3 - 5\\ &= 12 - 5\\ &= 7\end{align}

Thus, $$n = 7$$ and $$r = 3$$

## Chapter 8 Ex.8.2 Question 11

Prove that the coefficient of $$x^n$$ in the expansion of $$\left( {1 + x} \right)^{2n}$$ is twice the coefficient of $$x^n$$ in the expansion of$$\left( {1 - x} \right)^{2n - 1}$$.

### Solution

It is known that $$\left( {r + 1} \right)^{th}$$ term, $$\left( {T_{r + 1}} \right)$$ the binomial expression of $$\left( {a + b} \right)^n$$ is given by

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

Assuming that $$x^n$$ occurs in the $$\left( {r + 1} \right)^{th}$$ term of the expansion$$\left( {1 + x} \right)^{2n},$$ we obtain

\begin{align}{T_{r + 1}} &= {}^{2n}{C_r}{\left( 1 \right)^{2n - r}}{\left( x \right)^r}\\& = {}^{2n}{C_r}{x^r}\end{align}

Comparing the indices of $$x$$ in $$x^n$$ and in $$T_{r + 1},$$ we obtain $$r = n$$

Therefore, the coefficient of $$x^n$$ in the expansion of $$\left( {1 + x} \right)^{2n}$$

\begin{align}{}^{2n}{C_n} &= \frac{{\left( {2n} \right)!}}{{n!\left( {2n - n} \right)!}}\\ &= \frac{{\left( {2n} \right)!}}{\left( {n!} \right)\left( {n!} \right)}\\& = \frac{{\left( {2n} \right)!}}{\left( {n!} \right)^2} \qquad \quad \ldots \left( 1 \right)\end{align}

Assuming that $$x^n$$ occurs in the $$\left( {k + 1} \right)^{th}$$ term of the expansion $$\left( {1 + x} \right)^{2n - 1},$$ we obtain

\begin{align}T_{k + 1} &= {}^{2n}{C_k}{\left( 1 \right)^{2n - 1 - k}}{\left( x \right)^k}\\ &= {}^{2n}{C_k}{x^k}\end{align}

Comparing the indices of $$x$$ in $$x^n$$ and in $$T_{k + 1}$$, we obtain $$k = n$$

Therefore, the coefficient of $$x^n$$ in the expansion of $$\left( {1 + x} \right)^{2n - 1}$$

\begin{align}{}^{2n - 1}{C_n} &= \frac{{\left( {2n - 1} \right)!}}{{n!\left( {2n - 1 - n} \right)!}}\\ &= \frac{{\left( {2n - 1} \right)!}}{{n!\left( {n - 1} \right)!}}\\ &= \frac{{2n \cdot \left( {2n - 1} \right)!}}{{2n \cdot n!\left( {n - 1} \right)!}}\\& = \frac{{\left( {2n} \right)!}}{{2 \cdot \left( {n!} \right)\left( {n!} \right)}}\\ &= \frac{1}{2}\left[ {\frac{{\left( {2n} \right)!}}{{\left( {n!} \right)}^2}} \right] \qquad \quad \ldots \left( 2 \right)\end{align}

From (1) and (2), we can observe that

\begin{align}{}^{2n - 1}{C_n} &= \frac{1}{2}\left( {}^{2n}{C_n} \right)\\{}^{2n}{C_n} &= 2\left( {}^{2n - 1}{C_n} \right)\end{align}

Therefore, the coefficient of $$x^n$$ in the expansion of $$\left( {1 + x} \right)^{2n}$$ is twice the coefficient of $$x^n$$ in the expansion of$$\left( {1 - x} \right)^{2n - 1}.$$

Hence Proved

## Chapter 8 Ex.8.2 Question 12

Find a positive value of $$m$$ for which the coefficient of $$x^2$$ in the expansion $$\left( {1 + x} \right)^m$$ is $$6.$$

### Solution

It is known that $$\left( {r + 1} \right)^{th}$$ term, $$\left( {T_{r + 1}} \right)$$ the binomial expression of $$\left( {a + b} \right)^n$$ is given by

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

Assuming that $$x^2$$ occurs in the $$\left( {r + 1} \right)^{th}$$ term of the expansion $$\left( {1 + x} \right)^m,$$ we obtain

$T_{r + 1} = {}^m{C_r}{\left( 1 \right)^{m - r}}\left( x \right)^r = {}^m{C_r}\left( x \right)^r$

Comparing the indices of $$x$$ in $$x^2$$ in $$\left({T_{r + 1}} \right)$$, we obtain $$r = 2$$

Therefore, the coefficient of $${x^2}$$is $${}^m{C_2}$$.

\begin{align}{}^m{C_2} &= 6\\\frac{{m!}}{{2!\left( {m - 2} \right)!}} &= 6\\\frac{{m\left( {m - 1} \right) \times \left( {m - 2} \right)!}}{{2 \times \left( {m - 2} \right)!}} &= 6\\m\left( {m - 1} \right) &= 12\\{m^2} - m - 12 &= 0\\{m^2} - 4m + 3m - 12 &= 0\\m\left( {m - 4} \right) + 3\left( {m - 4} \right) &= 0\\\left( {m - 4} \right)\left( {m + 3} \right) &= 0\end{align}

$$m = 4$$ or $$m = - 3$$

Thus, $$4$$ is the positive value of $$m$$ for which the coefficient of $${x^2}$$ in the expansion $${\left( {1 + x} \right)^m}$$ is $$6.$$

Instant doubt clearing with Cuemath Advanced Math Program