Question

Suppose $l,m,n$ respectively represent the coefficient of $x^{10}$, the constant term and the coefficient of $x^{-10}$ in the expansion of ${\left(a{x}^2+\frac{b}{x^3}\right)}^{15}$.If $\frac{l}{m}+\frac{m}{n}=\frac{26}{11}$, then $a^2:b^2=$

• A. $16:9$
• B. $9:4$
• C. $4:1$
• D. $1:25$

Answer 1

Answer: (B)

Given binomial ${\left(a{x}^2+\frac{b}{x^3}\right)}^{15}$,
so the general term, $T_{r+1}={}^{15}{C}_r{a}^{15-r}{b}^r{x}^{30-5r}$
$\therefore$ The coefficient of $x^{10}$ is ( at $r=4$ )
$={}^{15}{C}_u{a}^{11}{b}^4=$ (given)
Similarly, $m=$ the constant term (at $r=6$ )
$={}^{15}{C}_6{a}^9{b}^6$
and $n=$ coefficient of $x^{-10}$ is $($ at $r=8)={}^{15}{C}_8{a}^7{b}^8$
$\because \frac{l}{m}+\frac{m}{n}=\frac{26}{11}$
$\Rightarrow \frac{{}^{15}{C}_4{a}^{11}{b}^4}{{}^{15}{C}_6{a}^9{b}^6}+\frac{{}^{15}{C}_6{a}^9{b}^6}{{}^{15}{C}_8{a}^7{b}^8}=\frac{26}{11}$
$\Rightarrow \frac{6!9!a^2}{4!111b^2}+\frac{8!7!a^2}{6!9!b^2}=\frac{26}{11}$
$\Rightarrow \frac{6\times 5}{11\times 10}\left(\frac{a^2}{b^2}\right)+\frac{7}{9}\left(\frac{a^2}{b^2}\right)=\frac{26}{11}$
$\Rightarrow \frac{a^2}{b^2}\left(\frac{3}{11}+\frac{7}{9}\right)=\frac{26}{11}$
$\Rightarrow \frac{a^2}{b^2}\left(\frac{27+77}{99}\right)=\frac{26}{11}$
$\Rightarrow \frac{a^2}{b^2}=\frac{9\times 26}{104}=\frac{9}{4}$