Question

The sum of the series $2C_0+\frac{C_1}{2}\cdot 2^2+\frac{C_2}{3}\cdot 2^3+\frac{C_3}{4}\cdot 2^4+\cdots +\frac{C_n}{n+1}\cdot 2^{n+1}$ is equal to , where$\left(C_r{=}^n{C}_r\right)$

• A. $\frac{3^{n+1}-1}{n-1}$
• B. $\frac{3^{n+1}-1}{n+1}$
• C. $\frac{3^{n+1}+1}{n+1}$
• D. $\frac{3^{n-1}-1}{n+1}$

Answer 1

The correct option is B $\frac{3^{n+1}-1}{n+1}$

We have,
$2C_0+\frac{C_1}{2}\cdot 2^2+\frac{C_2}{3}\cdot 2^3+\frac{C_3}{4}\cdot 2^4+\cdots +\frac{C_n}{n+1}\cdot 2^{n+1}$
$=\sum _{r=0}^n\frac{1}{r+1}{}^n{C}_r\cdot 2^{r+1}$
$=\frac{1}{n+1}\sum _{r=0}^n\frac{n+1}{r+1}{}^n{C}_r\cdot 2^{r+1}$
$=\frac{1}{n+1}\sum _{r=0}^n{}^{n+1}{C}_{r+1}\cdot 2^{r+1}$

$=\frac{1}{n+1}\sum _{s=1}^{n+1}{}^{n+1}{C}_s\cdot 2^s[\text{Putting}r+1=s]$

$=\frac{1}{n+1}[\left(\sum _{s=0}^{n+1}{}^{n+1}{C}_s{2}^s\right)-{}^{n+1}{C}_0\cdot 2^0]$
$=\frac{1}{n+1}\left[\right(1+2{)}^{n+1}-1]=\frac{3^{n+1}-1}{n+1}$