The sum of $(n + 1)$  terms of the series

$\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\frac{C_3}{5}+\dots$ is

We have 

$\begin{array}{l}\Rightarrow (1-x{)}^n=C_0-C_{1}x+C_{2}x-C_{3}x+\dots +(-1{)}^n{C}_n{x}^n\\ \\ \Rightarrow {\int }_0^1 x(1-x{)}^{n}dx={\int }_0^1 \left[C_{0}x-C_1{x}^2+C_2{x}^3-C_3{x}^4+\dots \right.\\ \left.+(-1{)}^n{C}_n{x}^{n+1}\right]dx \underset{ }{\to \left(1\right)}\end{array}$

L.H.S. of $\left(1\right)$ 

$\begin{array}{l}={\int }_0^1 x(1-x{)}^{n}dx={\int }_0^1 (1-x\left)\right(1-(1-x){)}^{n}dx\\ \left.{\left.={\int }_0^1 (1-x)x^{n}dx=\left(\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}\right)\right]}_0^{1}f\left(x\right)dx={\int }_0^1 f(a-x)dx\right]\\ =\frac{1}{(n+1)(n+2)}\end{array}$

R.H.S of (1)  

$=\left(C_0\frac{x^2}{2}-C_1\frac{x^3}{3}+C_2\frac{x^4}{4}-\right.{\left.\left.\dots +\frac{(-1{)}^n{C}_n}{n+1}{x}^{n+1}\right)\right]}_0^1$

$=\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\dots +\frac{(-1{)}^n{C}_n}{n+1}$

Thus,       $\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\dots +(-1{)}^n\frac{C_n}{n+1}$

$=\frac{1}{(n+1)(n+2)}$