Question

In the expansion of ${(x^2-\frac{1}{4})}^n$, the coefficient of third term is $31$, then the value of $n$ is-

• A. $30$
• B. $31$
• C. $29$
• D. $32$

Answer 1

Answer: (D)

$32$

We know that $T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

$T_3{=}^n{C}_2{\left(x^2\right)}^{n-2}{\left(\frac{-1}{4}\right)}^2$

$=\frac{n(n-1)}{2}{x}^{2n-4}\times \frac{1}{16}$

$=\frac{n(n-1)}{32}{x}^{2n-4}$

Coefficient of $T_3=\frac{n(n-1)}{32}=31$

$\Rightarrow n(n-1)=32(32-1)$

$\Rightarrow n=32$