Question

The value of $2\left({}^n{C}_0\right)+\frac{3}{2}\left({}^n{C}_1\right)+\frac{4}{3}\left({}^n{C}_2\right)+\frac{5}{4}\left({}^n{C}_3\right)...$ 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 $\frac{2^n(n+3)-1}{n+1}$

${(1+x)}^n{=}^{}{}^n{C}_0{+}^{}{}^n{C}_{1}x{+}^{}{}^n{C}_2{x}^2+...{+}^{}{}^n{C}_n{x}^n$

Integrating wrt x between limits $0$ and $x$ we get,

$\frac{{(1+x)}^{n+1}}{n+1}{=}^{}{}^n{C}_{0}x{+}^{}{}^n{C}_1\frac{x^2}{2}{+}^{}{}^n{C}_2\frac{x^3}{3}+....+\frac{x^{n+1}}{n+1}+..$

$\frac{{(1+x)}^{n+1}-1}{n+1}{=}^{}{}^n{C}_{0}x{+}^{}{}^n{C}_1\frac{x^2}{2}{+}^{}{}^n{C}_2\frac{x^3}{3}+....$

Multiply with $x$ and then differentiate to get

$\Rightarrow \frac{d\left(x\left(\frac{{(1+x)}^{n+1}-1}{n+1}\right)\right)}{dx}=\frac{d\left({}^{}{}^n{C}_0{x}^2{+}^{}{}^n{C}_1\frac{x^3}{2}+.....\right)}{dx}$

$\Rightarrow \frac{{(1+x)}^{n+1}+x(n+1){(1+x)}^n-1}{n+1}=2^{}{}^n{C}_{0}x+{\frac{3}{2}}^{}{}^n{C}_1{x}^2+{\frac{4}{3}}^{}{}^n{C}_2{x}^3+...$

Put $x=1$ to get,

$\Rightarrow \frac{{\left(2\right)}^{n+1}+(n+1){\left(2\right)}^n-1}{n+1}=2^{}{}^n{C}_0+{\frac{3}{2}}^{}{}^n{C}_1+{\frac{4}{3}}^{}{}^n{C}_2+...=\frac{2^n(n+3)-1}{n+1}$