Question

$\sum _{k=0}^n\frac{{}^{C}k}{k+1}=\frac{2^{n+1}-1}{n+1}$

Answer 1

$\sum _{k=0}^n\frac{C_k}{k+1}=\frac{2^{n+1}-1}{n+1}$

We know that :

${(1+x)}^n{=}^n{C}_0{+}^n{C}_1{x}^1{+}^n{C}_2{x}^2{+}^n{C}_3{x}^3+...{+}^n{C}_n{x}^n....1$

Integrating equation $1$ between $0$ ad $1$ w.r.t. $x$

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

${\left[\frac{{(1+x)}^{n+1}}{n+1}\right]}_0^1{=}^n{C}_0+\frac{{}^n{C}_1}{2}+\frac{{}^n{C}_2}{3}+...+\frac{{}^n{C}_n}{n+1}-0$

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

i.e $\left[\sum _{k=0}^n\frac{{}^n{C}_k}{k+1}\right]=\frac{2^{n+1}}{n+1}-1$

Hence proved.

*Tip: Always go for basic expand.

${(1+x)}^n,{(1+x)}^{-n},{(1-x)}^n,{(1-x)}^{-n}$