Question

$1+\frac{1}{4}+\frac{1\times 3}{4\times 8}+\frac{1\times 3\times 5}{4\times 8\times 12}+\cdots =$

• A. $\sqrt{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\sqrt{3}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

Answer: (A)

Let the given series be identical with
$(1+x{)}^n=1+nx+\frac{n(n-1)}{1\times 2}{x}^2+\cdots \infty$
or $nx=\frac{1}{4}$ or $n^2{x}^2=\frac{1}{16}$
Also, $\frac{n(n-1)}{2}{x}^2=\frac{3}{32}$ or $\frac{2n}{n-1}=\frac{\frac{1}{16}}{\frac{3}{32}}=\frac{2}{3}$
or $3n=n-1$
or$2n=-1$
or $n=-\frac{1}{2}$
$\Rightarrow x=-\frac{1}{2}$
$\therefore$ Required sum$={\left(1-\frac{1}{2}\right)}^{-\frac{1}{2}}={\left(\frac{1}{2}\right)}^{-\frac{1}{2}}$
$=(2{)}^{\frac{1}{2}}=\sqrt{2}$