Question

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

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

Answer 1

Answer: (D)

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