Question

Sum upto $n$ terms for series $\frac{C_0}{1\cdot 2}+\frac{C_1}{2\cdot 3}+\frac{C_2}{3\cdot 4}+\cdots$ is
( where $C_r={}^n{C}_r$)

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

Answer 1

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

Let $S=\frac{C_0}{1\cdot 2}+\frac{C_1}{2\cdot 3}+\frac{C_2}{3\cdot 4}+\cdots$ upto $n$ terms
$=\sum _{r=0}^n\frac{{}^n{C}_r}{(r+1)(r+2)}$
$=\sum _{r=0}^n\frac{{}^{n+2}{C}_{r+2}}{(n+1)(n+2)}[\because \frac{{}^{n+2}{C}_{r+2}}{{}^n{C}_r}=\frac{(n+2)(n+1)}{(r+2)(r+1)}]$
$=\frac{1}{(n+1)(n+2)}\sum _{r=0}^n{}^{n+2}{C}_{r+2}$
$=\frac{1}{(n+1)(n+2)}[{}^{n+2}{C}_2+{}^{n+2}{C}_3+{}^{n+2}{C}_4+\cdots +{}^{n+2}{C}_{n+2}]$
$=\frac{1}{(n+1)(n+2)}[2^{n+2}-{}^{n+2}{C}_0-{}^{n+2}{C}_1]$
$=\frac{2^{n+2}-n-3}{(n+1)(n+2)}$