Question

The value of $2{(}^n{C}_0)+\frac{3}{2}{(}^n{C}_1)+\frac{4}{3}{(}^n{C}_2)+\frac{5}{4}{(}^n{C}_3)+\cdots$ is

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

Answer 1

The correct option is B

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

$(1+x{)}^n{=}^n{C}_0{+}^n{C}_{1}x{+}^n{C}_2{x}^2{+}^n{C}_3{x}^3{\cdots }^n{C}_n{x}^n$ on Integrating from 0 to 1
$\frac{(1+x{)}^{n+1}-1}{n+1}{=}^n{C}_{0}x+\frac{{}^n{C}_1{x}^2}{2}+\frac{{}^n{C}_2{x}^3}{3}+\frac{{}^n{c}_3{x}^4}{4}\cdots$
Multiplying with x & differentiating
$\frac{d}{dx}\left\{x\left(\frac{(1+x{)}^{n+1}-1}{n+1}\right)\right\}=\frac{d}{dx}\{{}^n{C}_0{x}^2+\frac{{}^n{C}_1{x}^3}{2}+\frac{{}^n{C}_2{x}^4}{3}+\frac{{}^n{C}_3{x}^5}{5}\cdots \}\frac{(1+x{)}^{n+1}+x(n+1\left)\right(1+x{)}^n-1}{n+1}=2^n{C}_{0}x+\frac{3^n{C}_1{x}^2}{2}+\frac{4^n{C}_2{x}^3}{3}\cdots$
put x=1
$\frac{2^{n+1}+(n+1)2^n-1}{n+1}=2^n{C}_0+\frac{3}{2}{}^n{C}_1+\frac{4}{3}{}^n{C}_2+\cdots =\frac{2^n(n+3)-1}{n+1}$