Question

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

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

Answer 1

Answer: (C)

$S=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{8}\right)+\dots ..\text{ up to }n\text{ terms }=n-\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{2^2}+\dots .\text{ up to }n\text{ terms }\right)$