Question

Sum of the first n terms of the series
$\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\cdots$ is equal to

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

Answer 1

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

$LetS=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\cdots nterms=(1-\frac{1}{2})+(1-\frac{1}{4})+(1-\frac{1}{8})+(1-\frac{1}{16})\cdots nterms=(1+1+1+\cdots nterms)-(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\cdots +\frac{1}{2^n})=n-\left[\frac{\frac{1}{2}(1-\frac{1}{2^n})}{1-\frac{1}{2}}\right]=n-1+2^{-n}$