Question

If $(1+x{)}^n=\sum _{r=0}^n{C}_r{x}^r$, then the value of
$\frac{2^2{C}_0}{1.2}+\frac{2^3{C}_1}{2.3}+\frac{2^4{C}_2}{3.4}+\cdots +\frac{2^{n+2}{C}_n}{(n+1)(n+2)}$ equals

• A. $\frac{3^{n+2}-2n-5}{(n+1)(n+2)}$
• B. $\frac{3^n-2n-5}{n(n+2)}$
• C. $\frac{3^{n+1}-2n-5}{(n+1)(n+2)}$
• D. none of these

Answer 1

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

Here, $T_r=\frac{2^{r+1}{C}_{r-1}}{r(r+1)}=\frac{2^{r+1}(n+2)(n+1)}{(n+2)(n+1)\left(r\right)(r+1)}{}^n{C}_{r-1}$
$\therefore S=\sum _{r=1}^{n+1}{T}_r=\sum _{r=1}^{n+1}\frac{1}{(n+2)(n+1)}\left(\frac{n+2}{r+1}\right)\left(\frac{n+1}{r}\right){}^n{C}_{r-1}\times 2^{r+1}$
$=\sum _{r=1}^{n+1}\left(\frac{{1.2}^{r+1}}{(n+2)(n+1)}{}^{n+2}{C}_{r+1}\right)$
$=\frac{1}{(n+2)(n+1)}[{}^{n+2}{C}_2{2}^2{+}^{n+2}{C}_3{2}^3+\cdots {+}^{n+2}{C}_{n+2}{2}^{n+2}]$
$=\frac{1}{(n+2)(n+1)}[({}^{n+2}{C}_0+{}^{n+2}{C}_{1}2+{}^{n+2}{C}_2{2}^2+{}^{n+2}{C}_3{2}^3+\cdots +{}^{n+2}{C}_{n+2}{2}^{n+2})-{}^{n+2}{C}_0-{}^{n+2}{C}_{1}2]$
$=\frac{1}{(n+2)(n+1)}\left[\right(1+2{)}^{n+2}-2(n+2)-1]$
$=\frac{1}{(n+2)(n+1)}[3^{n+2}-2n-5]$