NCERT Solutions for Exercise 8.2 Class 11 Maths Chapter 8 - Binomial Theorem

Question:2 Find the coefficient of$a^5b^7$ in $(a- 2b)^{12}$

Answer:

As we know that the $(r+1)^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b)^n$ is given by

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

Now let's assume$a^5b^7$ happens in the $(r+1)^{th}$ term of the binomial expansion of $(a- 2b)^{12}$

So,

$T_{r+1}=^{12}C_rx^{12-r}(-2b)^r$

On comparing the indices of x we get,

$r=7$

Hence the coefficient of the $a^5b^7$ in $(a- 2b)^{12}$ is

$\\ \Rightarrow ^{12}C_7\times(-2)^7=\frac{12!}{5!7!}\times (-128)\\=\frac{12\times11\times10\times 9\times8}{5\times4\times3\times2}\times(-128) \\=-(729)(128) \\=-101376$

Question:3 Write the general term in the expansion of

$(x^2 - y)^6$

Answer:

As we know that the general $(r+1)^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b)^n$ is given by

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

So the general term of the expansion of $(x^2 - y)^6$ :

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

Question:4 Write the general term in the expansion of

$(x^2 - xy)^{12}, \ x\neq 0$

Answer:

As we know that the general $(r+1)^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b)^n$ is given by

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

So the general term of the expansion of $(x^2 - xy)^{12},$is

$\\\Rightarrow T_{r+1}\\=^{12}C_r(x^2)^{12-r}(-xy)^r\\=(-1)^r\times^{12}C_rx^{24-2r+r}y^r\\=(-1)^r\times^{12}C_rx^{24-r}y^r$.

Question:5 Find the 4^th term in the expansion of $(x-2y)^{12}$.

Answer:

As we know that the general $(r+1)^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b)^n$ is given by

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

So the $4^{th}$ term of the expansion of $(x-2y)^{12}$ is

$\\\Rightarrow T_4= T_{3+1}\\=^{12}C_3(x)^{12-3}(-2y)^3\\=(-2)^3\times^{12}C_3x^{9}y^3\\=-8\times\frac{12!}{3!9!}x^{9}y^3$

$\\=-8\times \frac{12\times11\times10}{3\times2}\times x^9y^3\\=-8\times220\times x^9y^3 \\=-1760x^9y^3$.

Question:6 Find the 13^th term in the expansion of $\left(9x - \frac{1}{3\sqrt x} \right )^{18},\ x\neq 0$

Answer:

As we know that the general $(r+1)^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b)^n$ is given by

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

So the $13^{th}$ term of the expansion of $\left(9x - \frac{1}{3\sqrt x} \right )^{18}$ is

$\\\Rightarrow T_{13}= T_{12+1}\\=^{18}C_{12}(9x)^{18-12}\left(\frac{1}{3\sqrt{x}}\right)^{12}\\ \\ \\=\frac{18!}{12!6!}\times9^{6}\left ( \frac{1}{3} \right )^{12}\times x^{6-6}$

$\\=\frac{18\times17\times16\times15\times14\times13}{6\times5\times4\times3\times2}\times9^{6}\left ( \frac{1}{3^{12}} \right )$

$=18564$

Question:7 Find the middle terms in the expansion of$\left(3 - \frac{x^3}{6} \right )^7$

Answer:

As we know that the middle terms in the expansion of $(a+b)^n$ when n is odd are,

$\left ( \frac{n+1}{2} \right )^{th}\:term\: \:and\:\:\left ( \frac{n+1}{2}+1 \right )^{th}\:term$

Hence the middle term of the expansion $\left(3 - \frac{x^3}{6} \right )^7$ are

$\left ( \frac{7+1}{2} \right )^{th}\:term\: \:and\:\:\left ( \frac{7+1}{2}+1 \right )^{th}\:term$

Which are $4^{th}\:term\:and\:\:5^{th} \:term$

Now,

As we know that the general $(r+1)^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b)^n$ is given by

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

So the $4^{th}$ term of the expansion of $\left(3 - \frac{x^3}{6} \right )^7$is

$\\\Rightarrow T_4= T_{3+1}\\=^{7}C_3(3)^{7-3}\left (- \frac{x^3}{6} \right )^3\\=\left ( -\frac{1}{6} \right )^3\times 3^4\times^{7}C_3\times x^{9}\\=\left ( -\frac{1}{6} \right )^3\times 3^4\times\frac{7!}{3!4!}\times x^{9} \\=-\frac{3\times3\times3\times3}{6\times6\times6}\times\frac{7\times 6\times5}{3\times2}\times x^9$

$=-\frac{105}{8}x^9$

And the $5^{th}$ Term of the expansion of $\left(3 - \frac{x^3}{6} \right )^7$is,

$\\\Rightarrow T_5= T_{4+1}\\=^{7}C_4(3)^{7-4}\left (- \frac{x^3}{6} \right )^4\\=\left ( -\frac{1}{6} \right )^4\times 3^3\times^{7}C_4\times x^{12}\\=\left ( -\frac{1}{6} \right )^4\times 3^3\times\frac{7!}{3!4!}\times x^{12} \\=\frac{3\times3\times3}{6\times6\times6\times6}\times\frac{7\times 6\times5}{3\times2}\times x^{12}$

$=\frac{35}{48}x^{12}$

Hence the middle terms of the expansion of given expression are

$-\frac{105}{8}x^9\:and\:\frac{35}{48}x^{12}.$

Question:8 Find the middle terms in the expansion of$\left(\frac{x}{3} + 9y \right )^{10}$

Answer:

As we know that the middle term in the expansion of $(a+b)^n$ when n is even is,

$\left ( \frac{n}{2}+1 \right )^{th}\:term\:$,

Hence the middle term of the expansion $\left(\frac{x}{3} + 9y \right )^{10}$ is,

$\left ( \frac{10}{2}+1 \right )^{th}\:term\:$

Which is $6^{th}\:term$

Now,

As we know that the general $(r+1)^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b)^n$ is given by

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

So the $6^{th}$ term of the expansion of $\left(\frac{x}{3} + 9y \right )^{10}$is

$\\\Rightarrow T_6= T_{5+1}\\=^{10}C_5\left ( \frac{x}{3} \right )^{10-5}\left ( 9y \right )^5\\$

$=\left ( \frac{1}{3} \right )^5\times9^5\times^{10}C_5\times x^5y^5$

$=\left ( \frac{1}{3} \right )^5\times9^5\times\left ( \frac{10!}{5!5!} \right )\times x^5y^5$

$=\left ( \frac{1}{3^5} \right )\times9^5\times\left ( \frac{10\times9\times8\times7\times6}{5\times4\times3\times2} \right )\times x^5y^5$

$=61236x^5y^5$

Hence the middle term of the expansion of $\left(\frac{x}{3} + 9y \right )^{10}$ is nbsp; $61236x^5y^5$.

Question:9 In the expansion of $(1 + a)^{m+n}$ , prove that coefficients of $a^m$ and $a^n$ are equal

Answer:

As we know that the general $(r+1)^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b)^n$ is given by

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

So, the general $(r+1)^{th}$ term $T_{r+1}$ in the binomial expansion of $(1 + a)^{m+n}$ is given by

$T_{r+1}=^{m+n}C_r1^{m+n-r}a^r=^{m+n}C_ra^r$

Now, as we can see $a^m$ will come when $r=m$ and $a^n$ will come when $r=n$

So,

Coefficient of $a^m$ :

$K_{a^m}=^{m+n}C_m=\frac{(m+n)!}{m!n!}$

CoeficientCoefficient of $a^n$ :

$K_{a^n}=^{m+n}C_n=\frac{(m+n)!}{m!n!}$

As we can see $K_{a^m}=K_{a^n}$.

Hence it is proved that the coefficients of $a^m$ and $a^n$ are equal.

Question:10 The coefficients of the $(r-1)$^th , r^th and $(r + 1)$^th terms in the expansion of $(x+1)^{n}$ are in the ratio 1 : 3 : 5. Find n and r.

Answer:

As we know that the general $(r+1)^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b)^n$ is given by

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

So,

$(r+1)^{th}$ Term in the expansion of $(x+1)^{n}$:

$T_{r+1}=^nC_rx^{n-r}1^r=^nC_rx^{n-r}$

$r^{th}$ Term in the expansion of $(x+1)^{n}$:

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

$(r-1)^{th}$ Term in the expansion of $(x+1)^{n}$:

$T_{r-1}=^nC_{r-2}x^{n-r+2}1^{r-2}=^nC_{r-2}x^{n-r+2}$

Now, As given in the question,

$T_{r-1}:T_r:T_{r+1}=1:3:5$

$^nC_{r-2}:^nC_{r-1}:^nC_{r}=1:3:5$

$\frac{n!}{(r-2)!(n-r+2)!}:\frac{n!}{(r-1)!(n-r+1)!}:\frac{n!}{r!(n-r)!}=1:3:5$

From here, we get ,

$\frac{r-1}{n-r+2}=\frac{1}{3}\:\:and\:\:\frac{r}{n-r+1}=\frac{3}{5}$

Which can be written as

$n-4r+5=0\:\:and\:\:3n-8r+3=0$

From these equations we get,

$n=7\:\:and\:\:r=3$

Question:11 Prove that the coefficient of $x^n$ in the expansion of $(1+x)^{2n}$ is twice the coefficient of $x^n$ in the expansion of $(1+x)^{2n-1}$.

Answer:

As we know that the general $(r+1)^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b)^n$ is given by

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

So, general $(r+1)^{th}$ term $T_{r+1}$ in the binomial expansion of $(1+x)^{2n}$ is,

$T_{r+1}=^{2n}C_r1^{2n-r}x^r$

$x^n$ will come when $r=n$,

So, Coefficient of $x^n$ in the binomial expansion of $(1+x)^{2n}$ is,

$K_{1x^n}=^{2n}C_n$

Now,

the general $(r+1)^{th}$ term $T_{r+1}$ in the binomial expansion of $(1+x)^{2n-1}$ is,

$T_{r+1}=^{2n-1}C_r1^{2n-1-r}x^r$

Here also $x^n$ will come when $r=n$,

So, Coefficient of $x^n$ in the binomial expansion of $(1+x)^{2n-1}$ is,

$K_{2x^n}=^{2n-1}C_n$

Now, As we can see

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

$^{2n-1}C_n=\frac{1}{2}\times^{2n}C_n$

$2\times^{2n-1}C_n=^{2n}C_n$

$2\times K_{2x^n}=K_{1x^n}$

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

Question:12 Find a positive value of m for which the coefficient of $x^2$ in the expansion $(1 + x)^ m$ is 6.

Answer:

As we know that the general $(r+1)^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b)^n$ is given by

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

So, the general $(r+1)^{th}$ term $T_{r+1}$ in the binomial expansion of $(1 + x)^ m$ is

$T_{r+1}=^mC_r1^{m-r}x^r=^mC_rx^r$

$x^2$ will come when $r=2$. So,

The coeficient of $x^2$ in the binomial expansion of $(1 + x)^ m$ = 6

$\Rightarrow ^mC_2=6$

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

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

$\Rightarrow m(m-1)=12$

$\Rightarrow m^2-m-12=0$

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

$\Rightarrow m=4\:or\:-3$

Hence the positive value of m for which the coefficient of $x^2$ in the expansion $(1 + x)^ m$ is 6, is 4.