Find the coefficient of:
Question:

Find the coefficient of:

(i) $x^{10}$ in the expansion of $\left(2 x^{2}-\frac{1}{x}\right)^{20}$

(ii) $x^{7}$ in the expansion of $\left(x-\frac{1}{x^{2}}\right)^{40}$

(iii) $x^{-15}$ in the expansion of $\left(3 x^{2}-\frac{a}{3 x^{3}}\right)^{10}$

(iv) $x^{9}$ in the expansion of $\left(x^{2}-\frac{1}{3 x}\right)^{9}$

(v) $x^{m}$ in the expansion of $\left(x+\frac{1}{x}\right)^{n}$

(vi) $x$ in the expansion of $\left(1-2 x^{3}+3 x^{5}\right)\left(1+\frac{1}{x}\right)^{8}$.

(vii) $a^{5} b^{7}$ in the expansion of $(a-2 b)^{12}$.

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

Solution:

(i) Suppose $x^{10}$ occurs in the $(r+1)$ th term in the given expression.

Then, we have:

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

Here,

$T_{r+1}={ }^{20} C_{r}\left(2 x^{2}\right)^{20-r}\left(\frac{-1}{x}\right)^{r}$

$=(-1)^{r}{ }^{20} C_{r}\left(2^{20-r}\right)\left(x^{40-2 r-r}\right)$

For this term to contain $x^{10}$, we must have :

$40-3 r=10$

$\Rightarrow 3 r=30$

 

$\Rightarrow r=10$

$\therefore$ Coefficient of $x^{10}=(-1)^{10}{ }^{20} C_{10}\left(2^{20-10}\right)={ }^{20} C_{10}\left(2^{10}\right)$

(ii) Suppose $x^{7}$ occurs at the $(r+1)$ th term in the given expression.

Then, we have:

$T_{r+1}={ }^{40} C_{r} x^{40-r}\left(\frac{-1}{x^{2}}\right)^{r}$

$=(-1)^{r}{ }^{40} C_{r} x^{40-r-2 r}$

For this term to contain $x^{7}$, we must have:

$40-3 r=7$

$\Rightarrow 3 r=40-7=33$

 

$\Rightarrow r=11$

$\therefore$ Coefficient of $x^{7}=(-1)^{11}{ }^{40} C_{11}=-{ }^{40} C_{11}$

(iii) Suppose $x^{-15}$ occurs at the $(r+1)$ th term in the given expression.

Then, we have:

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

$\Rightarrow T_{r+1}=(-1)^{r}{ }^{10} C_{r}\left(3^{10-r-r}\right)\left(x^{20-2 r-3 r}\right)\left(a^{r}\right)$

For this term to contain $x^{-15}$, we must have:

$20-5 r=-15$

$\Rightarrow 5 r=20+15$

$\Rightarrow r=7$

$\therefore$ Coef ficient of $x^{-15}=(-1)^{7}{ }^{10} C_{7} 3^{10-14}\left(a^{7}\right)=-\frac{10 \times 9 \times 8}{3 \times 2 \times 9 \times 9} a^{7}=-\frac{40}{27} a^{7}$

(iv) Suppose x9 occurs at the (+ 1)th term in the above expression.

Then, we have:

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

$=(-1)^{r}{ }^{9} C_{r}\left(x^{18-2 r-r}\right)\left(\frac{1}{3^{r}}\right)$

For this term to contain $x^{9}$, we must have :

$18-3 r=9$

$\Rightarrow 3 r=9$

$\Rightarrow r=3$

$\therefore$ Coefficient of $x^{9}=(-1)^{3}{ }^{9} C_{3} \frac{1}{3^{3}}=-\frac{9 \times 8 \times 7}{2 \times 9 \times 9}=\frac{-28}{9}$

(v) Suppose xm occurs at the (+ 1)th term in the given expression.

Then, we have:

$T_{r+1}={ }^{n} C_{r} x^{n-r} \frac{1}{x^{r}}$

$={ }^{n} C_{r} x^{n-2 r}$

 

For this term to contain $x^{m}$, we must have:

$n-2 r=m$

$\Rightarrow r=(n-m) / 2$

$\therefore$ Coefficient of $x^{m}={ }^{n} C_{(n-m) / 2}=\frac{n !}{\left(\frac{n-m}{2}\right) !\left(\frac{n+m}{2}\right) !}$

(vi) Suppose x occurs at the (+ 1)th term in the given expression.

Then, we have:

$\left(1-2 x^{3}+3 x^{5}\right)\left(1+\frac{1}{x}\right)^{8}$

$=\left(1-2 x^{3}+3 x^{5}\right)\left({ }^{8} C_{0}+{ }^{8} C_{1}\left(\frac{1}{x}\right)+{ }^{8} C_{2}\left(\frac{1}{x}\right)^{2}+{ }^{8} C_{3}\left(\frac{1}{x}\right)^{3}+{ }^{8} C_{4}\left(\frac{1}{x}\right)^{4}+{ }^{8} C_{5}\left(\frac{1}{x}\right)^{5}+{ }^{8} C_{6}\left(\frac{1}{x}\right)^{6}+{ }^{8} C_{7}\left(\frac{1}{x}\right)^{7}+{ }^{8} C_{8}\left(\frac{1}{x}\right)^{8}\right)$

$x$ occurs in the above expression at $-2 x^{3} \cdot{ }^{8} C_{2}\left(\frac{1}{x^{2}}\right)+3 x^{5} \cdot{ }^{8} C_{4}\left(\frac{1}{x}\right)^{4}$.

$\therefore$ Coefficient of $x=-2\left(\frac{8 !}{2 ! 6 !}\right)+3\left(\frac{8 !}{4 ! 4 !}\right)=-56+210=154$

(vii) Suppose a5 b7 occurs at the (r + 1)th term in the given expression.

Then, we have:

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

$=(-1)^{r}{ }^{12} C_{r}\left(a^{12-r}\right)\left(b^{r}\right)\left(2^{r}\right)$

For this term to contain $a^{5} b^{7}$, we must have :

$12-r=5$

$\Rightarrow r=7$

$\therefore$ Required coefficient $=(-1)^{7}{ }^{12} C_{7}\left(2^{7}\right)=-\frac{12 \times 11 \times 10 \times 9 \times 8 \times 128}{5 \times 4 \times 3 \times 2}=-101376$

(viii) Suppose x occurs at the (+ 1)th term in the given expression.

Then, we have:

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

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

$\left({ }^{16} C_{0}+{ }^{16} C_{1}(-x)+{ }^{16} C_{2}(-x)^{2}+{ }^{16} C_{3}(-x)^{3}+{ }^{16} C_{4}(-x)^{4}+{ }^{16} C_{5}(-x)^{5}+{ }^{16} C_{6}(-x)^{6}+{ }^{16} C_{7}(-x)^{7}+{ }^{16} C_{8}(-x)^{8}\right.$$\left.+{ }^{16} C_{9}(-x)^{9}+{ }^{16} C_{10}(-x)^{10}+{ }^{16} C_{11}(-x)^{11}+{ }^{16} C_{12}(-x)^{12}+{ }^{16} C_{13}(-x)^{13}+{ }^{16} C_{14}(-x)^{14}+{ }^{16} C_{15}(-x)^{15}+{ }^{16} C_{16}(-x)^{16}\right)$

$x$ occurs in the above expression at ${ }^{16} C_{1}(-x)-3 x^{16} C_{0}$.

$\therefore$ Coefficient of $x=-\left(\frac{16 !}{1 ! 15 !}\right)-3\left(\frac{16 !}{0 ! 16 !}\right)=-16-3=-19$

 

 

 

 

 

 

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