Question

If the sum of the coefficients of $x^{7}and{x}^4$ in the expansion of ${(\frac{x^2}{a}-\frac{b}{x})}^{11}$ is zero, then ab=

Answer 1

General term, $T_{r+1}{=}^{11}{C}_r{\left(\frac{x^2}{a}\right)}^r{\left(\frac{-b}{x}\right)}^{11-r}$

${=}^{11}{C}_r{a}^{-r}(-b{)}^{11-r}.x^{3r-11}$

$coeff.of\underset{–}{x^7}{=}^{11}{C}_7{a}^{-7}(-b{)}^4$..............(i)

$coeff.of\underset{–}{x^4}{=}^{11}{C}_4{a}^{-4}(-b{)}^7={-}^{11}{C}_4{a}^{-4}(b{)}^7$................(ii)

(i)$+$(ii)$=0$

${}^{11}{C}_7{a}^{-7}(-b{)}^4{-}^{11}{C}_4{a}^{-4}(b{)}^7=0$

${\Rightarrow }^{11}{C}_7{a}^{-7}(-b{)}^4{=}^{11}{C}_4{a}^{-4}(b{)}^7$

As ${}^{11}{C}_4{=}^{11}{C}_7\Rightarrow \frac{b^4}{a^7}=\frac{b^7}{a^4}$

$\Rightarrow a^3{b}^3=1\Rightarrow ab=1$.