Question

If ${(1+x)}^n=C_0+C_{1}x+C_2{x}^2+...+C_n{x}^n$, then $2C_0+2^2.\frac{C_1}{2}+2^3.\frac{C_2}{3}+...+2^{n+1}.\frac{C_n}{n+1}=$

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

Answer 1

Answer: (A)

$\frac{3^{n+1}-1}{n+1}$

We have,

$t_{r+1}=2^{r+1}\frac{{}^n{C}_r}{r+1}=2^{r+1}\frac{1}{n+1}{.}^{n+1}{C}_{r+1}$

Putting $r=0,1,2,...,n$ and adding, we get the required sum

$=\frac{1}{n+1}\{{2.}^{n+1}{C}_1+2^2{.}^{n+1}{C}_2+...+2^{n+1}{.}^{n+1}{C}_{r+1}\}$

$=\frac{1}{n+1}\left\{{(1+2)}^{n+1}{-}^{n+1}{C}_0\right\}=\frac{3^{n+1}-1}{n+1}$.