Question

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

Answer: (B)

We know that
$(1+x{)}^n={}^n{C}_0+x{}^n{C}_1+x^2{}^n{C}_2+\dots +x^n{}^n{C}_n$
On integrating both sides from 0 to 2 , we get
${\left[\frac{(1+x{)}^{n+1}}{n+1}\right]}_0^2$
$={\left[x^n{C}_0+\frac{x^2}{2}{}^n{C}_1+\frac{x^3}{3}{}^n{C}_2+\dots +\frac{x^{n+1}}{n+1}{}^n{C}_n\right]}_0^2$
$\Rightarrow \frac{(3{)}^{n+1}}{n+1}-\frac{1}{n+1}=2{}^n{C}_0+\frac{2^2}{2}{}^n{C}_1+\frac{2^3}{3}{}^n{C}_2+\dots +\frac{2^{n+1}}{n+1}{}^n{C}_n-0$
$\Rightarrow \frac{2}{1}{}^n{C}_0+\frac{2^2}{2}{}^n{C}_1+\frac{2^3}{3}{}^n{C}_2+\dots +\frac{2^{n+1}}{n+1}{}^n{C}_n$
$=\frac{3^{n+1}-1}{n+1}$