Question

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

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

Answer 1

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

Given $\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+...$

$=(1-\frac{1}{2})+(1-\frac{1}{4})+(1-\frac{1}{8})+(1-\frac{1}{16})+...$

$=n-(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+.........+\frac{1}{2^n})$

$=n-\frac{\frac{1}{2}(1-2^{-n})}{1-\frac{1}{2}}$ [Sum of GP $=\frac{a(1-r^n)}{1-r}$ when $r<1$]

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