Question

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

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

Answer 1

The correct option is C

$\mathrm{n}+2^{-\mathrm{n}}-1$

Explanation for the correct answer:

The given sum is $\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+......$

This can be simplified as

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

$=1+1+1+...+1(\mathrm{n}\mathrm{times})-\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{\mathrm{n}}}\right)$

$1+1+1+....+1(\mathrm{n}\mathrm{times})=\mathrm{n}$

$\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^n}\right)$ is the sum of a geometric progression where

first term of the G.P $a=\frac{1}{2}$ and common ratio $\text{r=}\frac{1}{2}$

Sum of $\mathrm{n}$ terms of a geometric progression when $\mathrm{r}<1$ is given as

${\mathrm{S}}_{\mathrm{n}}=\mathrm{a}\frac{\left(1-{\mathrm{r}}^{\mathrm{n}}\right)}{1-\mathrm{r}}$

$=\frac{1}{2}\times \frac{1-{\displaystyle \frac{1}{2^{\mathrm{n}}}}}{1-{\displaystyle \frac{1}{2}}}$

$\Rightarrow$ ${\mathrm{S}}_{\mathrm{n}}=1-\frac{1}{2^{\mathrm{n}}}$

Hence, the given sum can be calculated as

$\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+......=\mathrm{n}-{\mathrm{S}}_{\mathrm{n}}$

$=\mathrm{n}-1+\frac{1}{2^{\mathrm{n}}}$

$=\mathrm{n}-1+2^{-\mathrm{n}}$

Hence, the sum of the given series is $\mathrm{n}+2^{-\mathrm{n}}-1$.

Hence, option C is the correct answer.