Question

The sum of the series $ \frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+... $ upto $ n $ term is

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

Answer 1

Answer: (A)

$\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots$ 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
$=(1+1+\ldots n$ times $)$
$-\left[\frac{1}{2}+\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{3}+\ldots n\right.$ terms $]$
$=n-\frac{\frac{1}{2}\left[1-\left(\frac{1}{2}\right)^{n}\right]}{\left(1-\frac{1}{2}\right)}$
$=n-1+\frac{1}{2^{n}}$
Alternate Method :

$=\frac{1\left(2^{n}-1\right)}{2-1}=2^{n}-1$
$\therefore $ nth term of given series,
$T_{n}=\frac{2^{n}-1}{2^{n}}=1-\frac{1}{2^{n}}$
Required sum $=\sum T_{n}=\sum\left(1-\frac{1}{2^{n}}\right)$
$=\sum 1-\sum\left(\frac{1}{2^{n}}\right)$
$=n-\left(\frac{1}{2}+\frac{1}{2^{2}}+\ldots+\frac{1}{2^{n}}\right)$
$=n-\frac{\frac{1}{2}\left[1-\left(\frac{1}{2}\right)^{n}\right]}{\left(1-\frac{1}{2}\right)}$
$=n-1+\frac{1}{2^{n}}$