Question

The sum to $n$ terms of the series $\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\dots$ is

• A. $2^n-1$
• B. $1-2^n$
• C. $n+2^n-1$
• D. $n-1+2^{-n}$

Answer 1

Answer: (D)

$\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\dots .$ upto $n$ terms
$=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{8}\right)\dots$ upto $n$ terms
$=(1+1+1+\dots n$ terms $)-(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\dots n$ terms $)$
$=n-\frac{\frac{1}{2}\left(1-\frac{1}{2^n}\right)}{1-\frac{1}{2}}$
$=n-1+\frac{1}{2^n}=n-1+2^n$