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)

$n+\frac{4^{-n}}{3}-\frac{1}{3}$

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