Question

Find the coeffcient of:
i) $x^{-7}$ in the expansion of ${\left(ax-\frac{1}{b{x}^2}\right)}^8$
ii) $x^6$ in the expansion ${\left(a-b{x}^2\right)}^{10}$

Answer 1

i) In the expansion of $(ax-\frac{1}{b{x}^2}{)}^8$, the general term is $t_{r+1}={}^8{C}_r\times (ax{)}^{8-r}\times (\frac{-1}{b{x}^2}{)}^r$

$t_{r+1}={}^8{C}_r\times (a{)}^{8-r}\times (\frac{-1}{b}{)}^r\times x^{8-r-2r}$

Since, we need to find the coefficient of $x^{-7}$ ,so we equate it to the term of $x$ in the general term

$\because x^{-7}=x^{8-3r}\Rightarrow -7=8-3r\Rightarrow 3r=15\Rightarrow r=5\Rightarrow r+1=5+1=6$

Thus, the general term containing $x^{-7}$ is $6^{th}$ term of the expansion, i.e., $t_6={}^8{C}_5\times (ax{)}^{8-5}\times (\frac{-1}{b{x}^2}{)}^5$

$\Rightarrow t_6=\frac{8!}{(8-5)!5!}\times a^3{x}^3\times \frac{-1}{b^5{x}^{10}}$ $\Rightarrow t_6=-\frac{8\times 7\times 6}{3\times 2}\times \frac{a^3}{b^5}\times x^{-7}$$\Rightarrow t_6=-\frac{56a^3}{b^5}\times x^{-7}$

$\therefore -\frac{56a^3}{b^5}$ is the required coefficient.

ii) In the expansion of $(a-b{x}^2{)}^{10}$, the general term is $t_{r+1}={}^{10}{C}_r\times (a{)}^{10-r}\times {(-b{x}^2)}^r$

$t_{r+1}={}^{10}{C}_r\times a^{10-r}\times (-b{)}^r\times (x^2{)}^r$

Since, we need to find the coefficient of $x^6$ ,so we equate it to the term of $x$ in the general term

$\because x^6=x^{2r}\Rightarrow 6=2r\Rightarrow r=3\Rightarrow r+1=3+1=4$

Thus, the general term containing $x^6$ is $4^{th}$ term of the expansion, i.e., $t_4={}^{10}{C}_3\times (a{)}^{10-3}\times (-b{x}^2{)}^3$

$\Rightarrow t_4=-\frac{10!}{(10-3)!3!}\times a^7{b}^3\times x^6$ $\Rightarrow t_4=-\frac{10\times 9\times 8}{3\times 2}\times a^7{b}^3{x}^6$$\Rightarrow t_4=-120a^7{b}^3{x}^6$

$\therefore -120a^7{b}^3$ is the required coefficient.