Question

The sum of the series
$\frac{3}{4\cdot 8}-\frac{3\cdot 5}{4\cdot 8\cdot 12}+\frac{3\cdot 5\cdot 7}{4\cdot 8\cdot 12\cdot 16}-...$

• A. $\sqrt{\frac{3}{2}}-\frac{3}{4}$
• B. $\sqrt{\frac{2}{3}}-\frac{3}{4}$
• C. $\sqrt{\frac{3}{2}}-\frac{1}{4}$
• D. $\sqrt{\frac{2}{3}}-\frac{1}{4}$

Answer 1

Answer: (B)

Given series is
$\frac{3}{4\cdot 8}-\frac{3\cdot 5}{4\cdot 8\cdot 12}+\frac{3\cdot 5\cdot 7}{4\cdot 8\cdot 12\cdot 16}-\dots +\frac{3}{4}-\frac{3}{4}$
$=\frac{3}{4}+\frac{3}{4\cdot 8}-\frac{3\cdot 5}{4\cdot 8\cdot 12}+\frac{3\cdot 5\cdot 7}{4\cdot 8\cdot 12\cdot 16}-\dots -\frac{3}{4}$
$=1-\frac{1}{4}+\frac{1\cdot 3}{2\cdot 4\cdot 4}-\frac{1\cdot 3\cdot 5}{4\cdot 4\cdot 2\cdot 4\cdot 3}+\dots -\frac{3}{4}$
$=[1+1\left(-\frac{1}{4}\right)+\frac{1(1+2)}{2}{\left(-\frac{1}{4}\right)}^2$ $+\frac{1(1+2)(1+4)}{3\cdot 2}{\left(-\frac{1}{4}\right)}^3+\dots ]-\frac{3}{4}$
$=[1+\frac{1}{1!}\left(-\frac{1}{4}\right)+\frac{1(1+2)}{2!}{\left(-\frac{1}{4}\right)}^2+$
$\frac{1(1+2)(1+4)}{3!}{\left(-\frac{1}{4}\right)}^3+\dots .]-\frac{3}{4}$
$={\left(1-\frac{1}{4}\right)}^{-1/2}-\frac{3}{4}$
$=\sqrt{\frac{2}{3}}-\frac{3}{4}$