Question

If $(1+x{)}^n=\sum _{r=0}^n{}^n{C}_r{x}^n,$ then $\frac{C_0}{1\cdot 2}{2}^2+\frac{C_1}{2\cdot 3}{2}^3+\frac{C_2}{3\cdot 4}{2}^4+\cdots +\frac{C_n}{(n+1)(n+2)}{2}^{n+2}$ is equal to

• A. $\frac{3^{n+2}+2n-5}{(n+1)(n+2)}$
• B. $\frac{3^{n+2}-2n+5}{(n+1)(n+2)}$
• C. $\frac{3^{n+2}-2n-5}{(n+1)(n+2)}$
• D. $\frac{3^{n+2}+2n+5}{(n+1)(n+2)}$

Answer 1

The correct option is C $\frac{3^{n+2}-2n-5}{(n+1)(n+2)}$

$\frac{C_0}{1\cdot 2}{2}^2+\frac{C_1}{2\cdot 3}{2}^3+\frac{C_2}{3\cdot 4}{2}^4+\cdots +\frac{C_n}{(n+1)(n+2)}{2}^{n+2}$

$=\sum _{r=0}^n\frac{1}{(r+1)(r+2)}{}^n{C}_r\cdot 2^{r+2}$

$=\frac{1}{(n+1)(n+2)}\sum _{r=0}^n\frac{n+2}{r+2}\cdot \frac{n+1}{r+1}{}^n{C}_r{2}^{r+2}$

$=\frac{1}{(n+1)(n+2)}\sum _{r=0}^n{}^{n+2}{C}_{r+2}{2}^{r+2}$

Putting $r+2=s$
$=\frac{1}{(n+1)(n+2)}\left[\sum _{s=2}^{n+2}{}^{n+2}{C}_s{2}^s\right]$

$=\frac{1}{(n+1)(n+2)}[\left(\sum _{s=0}^{n+2}{}^{n+2}{C}_s{2}^s\right)-({}^{n+2}{C}_0{2}^0+{}^{n+2}{C}_1{2}^1\left)\right]$

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

$=\frac{3^{n+2}-2n-5}{(n+1)(n+2)}$