Question

Let $S=\frac{2}{1}{}^n{C}_0+\frac{2^2}{2}{}^n{C}_1+\frac{2^3}{3}{}^n{C}_2+\cdots +\frac{2^{n+1}}{n+1}{}^n{C}_n$. Then, $S$ equals

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

Answer 1

The correct option is B $\frac{3^{n+1}-1}{n+1}$

We know that
${(1+x)}^n{=}^n{C}_0+x{}^n{C}_1+x^2{}^n{C}_2+\cdots +x^n{}^n{C}_n$

On integrating both sides from $0$ to $2$, we get ${\left[\frac{{(1+x)}^{n+1}}{n+1}\right]}_0^2$

$={[x^n{C}_0+\frac{x^2}{2}{}^n{C}_1+\frac{x^3}{3}{}^n{C}_2+\cdots +\frac{x^{n+1}}{n+1}{}^n{C}_n]}_0^2$

$\Rightarrow \frac{{\left(3\right)}^{n+1}}{n+1}-\frac{1}{n+1}=2{}^n{C}_0+\frac{2^2}{2}{}^n{C}_1+\frac{2^3}{3}{}^n{C}_2+\cdots +\frac{2^{n+1}}{n+1}{}^n{C}_n-0$

$\Rightarrow \frac{2}{1}{}^n{c}_0+\frac{2^2}{2}{}^n{C}_1+\frac{2^3}{3}{}^n{C}_2+\cdots +\frac{2^{n+1}}{n+1}{}^n{C}_n=\frac{3^{n+1}-1}{n+1}$