Question

The value of $\frac{3}{2^3}+\frac{4}{2^4\cdot 3}+\frac{5}{2^6\cdot 3}+\frac{6}{2^7\cdot 5}+\cdots \infty$ is equal to

• A. $2$
• B. $\frac{1}{2}$
• C. $3$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $\frac{1}{2}$

$S=\sum _{r=1}^{\infty }\frac{r+2}{2^{r+1}r(r+1)}$
$=\sum _{r=1}^{\infty }\frac{2(r+1)-r}{2^{r+1}r(r+1)}$
$=\sum _{r=1}^{\infty }\frac{1}{2^{r+1}}(\frac{2}{r}-\frac{1}{r+1})$
$=\sum _{r=1}^{\infty }\frac{1}{2^r\cdot r}-\frac{1}{2^{r+1}(r+1)}$
$=(\frac{1}{2\cdot 1}-\frac{1}{2^2\cdot 2})+(\frac{1}{2^2\cdot 2}-\frac{1}{2^3\cdot 3})+\cdots \infty$
$=\frac{1}{2}$