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

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

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

Now, ${\int }_0^2{(1+x)}^n.dx={\int }_0^2({}^n{C}_0{+}^n{C}_{1}x{+}^n{C}_2{x}^2+...{+}^n{C}_n{x}^n).dx\Rightarrow {\left[\frac{{(1+x)}^{n+1}}{n+1}\right]}_0^2={\left[({}^n{C}_{0}x+\frac{{}^n{C}_1}{2}{x}^2+\frac{{}^n{C}_2}{3}{x}^3+...{+}^n{C}_n\frac{x^{n+1}}{n+1})\right]}_0^2=\frac{3^{n+1}-1}{n+1}=S$