Question

The value of $\frac{3}{4} + \frac{15}{16} +\frac{63}{64} +..... $ upto n terms is

• A. $n - \frac{4^n}{3} - \frac{1}{3}$
• B. $n + \frac{4^{-n}}{3} - \frac{1}{3}$
• C. $n + \frac{4^n}{3} - \frac{1}{3}$
• D. $n - \frac{4^{-n}}{3} + \frac{1}{3}$

Answer 1

Answer: (B)

Consider $\frac{3}{4}+\frac{15}{16}+\frac{63}{64}+...$ upto $n$ terms
$=\frac{2^{2}-1}{2^{2}}+\frac{2^{4}-1}{2^{4}}+\frac{2^{6}-1}{2^{6}}$ upto $n$ terms
$=\left(1-\frac{1}{2^{2}}\right)+\left(1-\frac{1}{2^{4}}\right)+\left(1-\frac{1}{2^{6}}\right)+...$ upto $n$ terms
$=(1+1+1+\ldots \text { upto } n \text { terms })-\left(\frac{1}{2^{2}}+\frac{1}{2^{4}}+\frac{1}{2^{6}}+\ldots \text { upto } n \text { terms }\right)$
$= n -\frac{1}{2^{2}}\left[\frac{1-\left(\frac{1}{2^{2}}\right)^{ n }}{1-\frac{1}{2^{2}}}\right]$
$= n +\frac{4^{- n }}{3}-\frac{1}{3}$