Question

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

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

Answer 1

The correct option is A $n+2^{-n}-1$

Sum of first $n$ terms,

$S=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+......n^{th}$terms

Breaking each term in $1-\frac{1}{2}$

$S=1-\frac{1}{2}+1-\frac{1}{4}+1-\frac{1}{8}+1-\frac{1}{16}+........=(1+1+\mathrm{1.....}nterms)-(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}.......nterms)=n-S_1-\left(i\right)S_1=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}........nterms$

Series is in $GP\left(a\right)=\frac{1}{2}$ and common difference $\left(r\right)=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{\frac{1}{8}}{\frac{1}{4}}=\frac{1}{2}$

$S_1=\frac{a(1-r^n)}{1-r}$ where$r<1$

$=\frac{\frac{1}{2}(1-{\left(\frac{1}{2}\right)}^n)}{1-\frac{1}{2}}=\frac{\frac{1}{2}(1-2^{-n})}{\frac{1}{2}}=1-2^{-n}$

So,putting value of $S_1$ in equation $\left(i\right)$

$S=n-(1-2^{-n})=n+2^{-n}-1$

Answer $\left(A\right)$