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)

$\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\dots$ 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
$=(1+1+\dots n$ times $)$
$-[\frac{1}{2}+{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^3+\dots n$ 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 $=\sum T_n=\sum \left(1-\frac{1}{2^n}\right)$
$=\sum 1-\sum \left(\frac{1}{2^n}\right)$
$=n-\left(\frac{1}{2}+\frac{1}{2^2}+\dots +\frac{1}{2^n}\right)$
$=n-\frac{\frac{1}{2}\left[1-{\left(\frac{1}{2}\right)}^n\right]}{\left(1-\frac{1}{2}\right)}$
$=n-1+\frac{1}{2^n}$