Question

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

Answer 1

The correct option is A

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

$\text{Let}S=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\cdots nterms.\text{i.e.,}S=(1-\frac{1}{2})+(1-\frac{1}{4})+(1-\frac{1}{8})+(1-\frac{1}{16})\cdots nterms=(1+1+1+\cdots nterms)-(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\cdots +\frac{1}{2^n})$
The sequence $\frac{1}{2},\frac{1}{2^2},\frac{1}{2^3},\frac{1}{2^4},\cdots ,\frac{1}{2^n}$ forms a GP with first term $a=\frac{1}{2}$ and common ratio $r=\frac{1}{2}.$
The sum to $n$ terms of a GP with first term $a$ and common ratio $r$ is
$S_n=\frac{a(1-r^n)}{1-r}.$
$\therefore S=n-\left[\frac{\frac{1}{2}(1-\frac{1}{2^n})}{1-\frac{1}{2}}\right]=n-1+2^{-n}$