Question

$1+\frac{2}{4}+\frac{2.5}{4.8}+\frac{\mathrm{2.5.8}}{\mathrm{4.8.12}}+\frac{\mathrm{2.5.8.11}}{\mathrm{4.8.12.16}}+...$ is equal to

• A. $4^{-2/3}$
• B. $\sqrt[3]{16}$
• C. $\sqrt[3]{4}$
• D. $4^{3/2}$

Answer 1

Answer: (B)

Let $S=1+\frac{2}{4}+\frac{2}{4}\cdot \frac{5}{8}+\frac{\mathrm{2.5.8}}{\mathrm{4.8.12}}+\dots$
On comparing with
$(1+x{)}^n=1+nx+\frac{n(n-1)}{2!}{x}^2+\dots .$,
we get $nx=\frac{2}{4}...(i)$
and $\frac{n(n-1)}{2!}{x}^2=\frac{2.5}{4.8}...(ii)$
From Eqs. (i) and (ii)
$\frac{\frac{n(n-1)}{2!}{x}^2}{n^2{x}^2}=\frac{\frac{2.5}{4.8}}{\frac{2.2}{4.2}}$
$\Rightarrow \frac{n-1}{n}=\frac{5}{2}$
$\Rightarrow 2n-2=5n$
$\Rightarrow n=-\frac{2}{3}$
On putting the value of $n$ in Eq. (i), we get
$-\frac{2}{3}x=\frac{2}{4}$
$\Rightarrow x=-\frac{3}{4}$
$\therefore S={\left(1+x^2\right)}^n={\left(1-\frac{3}{4}\right)}^{-2/3}={\left(\frac{1}{4}\right)}^{-2/3}=\sqrt[3]{16}$