Question

$1+\frac{3}{2}+\frac{5}{2^2}+\frac{7}{2^3}+...\infty isequalto$

• A. $3$
• B. $6$
• C. $9$
• D. $12$

Answer 1

The correct option is B

$6$

Explanation for the correct option:

Step 1: Simplify given series

Let $S=1+\frac{3}{2}+\frac{5}{2^2}+\frac{7}{2^2}+......\infty$ ………(1)

Multiply both side of equation (1) by $\frac{1}{2}$:

$\Rightarrow \frac{S}{2}=\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+......\infty$ ……….(2)

Subtract equation (2) from equation (1):

$\Rightarrow \frac{S}{2}=1+\frac{2}{2}+\frac{2}{2^2}+\frac{2}{3^2}+.......\infty$

$\Rightarrow \frac{S}{2}=1+\left(1+\frac{1}{2}+\frac{1}{2^2}+........\infty \right)$

Step 2: Use sum of infinite G.P. formula

This is a infinite G.P. with $a=1,r=\frac{1}{2}$

$\Rightarrow \frac{S}{2}=1+\left[\frac{1}{\left(1-{\displaystyle \frac{1}{2}}\right)}\right]$ $\left[Since,{\mathit{S}}_{\mathbf{\infty }}\mathbf{=}\frac{\mathbf{a}}{\mathbf{1}\mathbf{-}\mathbf{r}}\right]$

$\Rightarrow \frac{S}{2}=1+2=3$

$\Rightarrow S=\mathbf{6}$

Hence, Option ‘B’ is Correct.