Mathematics - Binomial Theorem Question with Solution | TestHub

Given,

The coefficient of $x^7$ in the expansion of  ${\left(a{x}^2-\frac{1}{bx}\right)}^{13}$ is equal to the coefficient of $x^{-5}$ in ${\left(ax+\frac{1}{b{x}^2}\right)}^{13}$.

We know that, the general term ${\mathrm{T}}_{r+1}$ in the expansion ${\left(a+b\right)}^n$  is

${\mathrm{T}}_{r+1}={}^{n}C_r{a}^{n-r}{b}^r$

Applying to ${\left(ax-\frac{1}{b{x}^2}\right)}^{13}$, we get

$T_{r+1}={}^{13}C_r{\left(ax\right)}^{13-r}{\left(-\frac{1}{b{x}^2}\right)}^r$

$\Rightarrow T_{r+1}={\left(-1\right)}^r\times {}^{13}C_r{\left(a\right)}^{13-r}{\left(x\right)}^{13-3r}{\left(b\right)}^{-r}$

Therefore, $13-3r=7\Rightarrow r=2$ for coefficient of ${\mathrm{x}}^7$.

Thus,

$T_3={}^{13}C_2\left(\frac{a^{11}}{b^2}\right)$

Similarly, applying to ${\left(ax+\frac{1}{b{x}^2}\right)}^{13}$, we get

$T_{r+1}={}^{13}C_r{\left(ax\right)}^{13-r}{\left(\frac{1}{b{x}^2}\right)}^r$

$\Rightarrow T_{r+1}={}^{13}C_r{\left(a\right)}^{13-r}{\left(x\right)}^{13-3r}{\left(b\right)}^{-r}$

Therefore, $13-3r=-5$ for coefficient of ${\mathrm{x}}^{-5}$

$\Rightarrow r=6$

So,

$T_7={}^{13}C_6{\left(a\right)}^7{\left(b\right)}^{-6}$

Hence, applying the given condition we get

${}^{13}C_2\left(\frac{a^{11}}{b^2}\right)={}^{13}C_6{\left(a\right)}^7{\left(b\right)}^{-6}$

$\Rightarrow a^4{b}^4=\frac{{}^{13}C_6}{{}^{13}C_2}$

$\Rightarrow a^4{b}^4=\frac{13!}{7!·6!}\times \frac{2!·11!}{13!}$

$\Rightarrow a^4{b}^4=\frac{11\times 10\times 9\times 8}{6\times 5\times 4\times 3}$

$\Rightarrow a^4{b}^4=22$