Question

If $C_0,C_1,C_2,......,C_n$ be the coefficients in the expansion of $(1+x{)}^n$, then $\frac{2^2.C_0}{1.2}+\frac{2^3.C_1}{2.3}+.....+\frac{2^{n+2}.C_n}{\left(n+1\right)\left(n+2\right)}$ is

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

Answer 1

Answer: (B)

We have,
$t_{r+1}=\frac{2^{r+2}{}^n{C}_r}{\left(r+1\right)\left(r+2\right)}=\frac{2^{r+2}}{r+2}.{\frac{1}{r+1}}^n{C}_r$
$=\frac{2^{r+2}}{r+2}.{\frac{1}{n+1}}^{n+1}{C}_{r+1}$
$=\frac{2^{r+2}}{r+2}.\left({\frac{1}{r+1}}^{n+1}{C}_{r+1}\right)$
$=\frac{2^{r+2}}{r+2}.{\frac{1}{n+2}}^{n+2}{C}_{r+2}$
$\left[\because {\frac{1}{r+1}}^n{C}_r={\frac{1}{r+1}}^{n+1}{C}_{r+1}\right]$
Putting $r=0,1,2,......,n$ and adding we get,
The given expression
$=\frac{1}{\left(n+1\right)\left(n+2\right)}\left\{2^2{.}^{n+2}{C}_2+2^3{.}^{n+2}{C}_3+....+2^{n+2}{.}^{n+2}{C}_{n+2}\right\}$
$=\frac{1}{\left(n+1\right)\left(n+2\right)}\left\{{\left(1+2\right)}^{n+2}{-}^{n+2}{C}_0-2{.}^{n+2}{C}_1\right\}$
$=\frac{3^{n+2}-2\left(n+2\right)-1}{\left(n+1\right)\left(n+2\right)}=\frac{3^{n+2}-2n-5}{\left(n+1\right)\left(n+2\right)}$