Question

$\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+....100$ terms $=$

• A. $2^{100}+99$
• B. $2^{-100}+99$
• C. $2^{-101}+99$
• D. $2^{-99}+99$

Answer 1

The correct option is B

$2^{-100}+99$

Explanation for the correct option:

Step1. Find the sum $\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\dots$

$=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{8}\right)+\dots$

$=\left(1+1+1+\dots \right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\dots \right)$

$=n-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...\right)$ $\left[\mathbf{\because }\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\mathbf{1}\mathbf{}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{}\mathbf{=}\mathbf{}\mathit{n}\right]$

The second series is a GP

Here, $a=\frac{1}{2},r=\frac{1}{2}$

Step 2. Apply the formula $\frac{\mathit{a}\mathbf{(}\mathbf{1}\mathbf{-}{\mathbf{r}}^{\mathbf{n}}\mathbf{)}\mathbf{}}{\mathbf{1}\mathbf{-}\mathbf{r}}$ in second series, we get

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

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

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

Put $n=100$

$=2^{-100}+100-1$

$={\mathbf{2}}^{\mathbf{-}\mathbf{100}}\mathbf{+}\mathbf{99}$

Hence, Option ‘B’ is Correct.