The sum of  (n + 1)  terms of the seriesC02−C13+C24−C35+… is

# The sum of   terms of the series$\frac{{C}_{0}}{2}-\frac{{C}_{1}}{3}+\frac{{C}_{2}}{4}-\frac{{C}_{3}}{5}+\dots$ is

1. A

2. B

$\frac{1}{n+2}$

3. C

$\frac{1}{n\left(n+1\right)}$

4. D

$\frac{1}{\left(n+1\right)\left(n+2\right)}$

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### Solution:

We have

L.H.S. of $\left(1\right)$

$\begin{array}{l}={\int }_{0}^{1} x\left(1-x{\right)}^{n}dx={\int }_{0}^{1} \left(1-x\right)\left(1-\left(1-x\right){\right)}^{n}dx\\ {={\int }_{0}^{1} \left(1-x\right){x}^{n}dx=\left(\frac{{x}^{n+1}}{n+1}-\frac{{x}^{n+2}}{n+2}\right)]}_{0}^{1}f\left(x\right)dx={\int }_{0}^{1} f\left(a-x\right)dx]\\ =\frac{1}{\left(n+1\right)\left(n+2\right)}\end{array}$

R.H.S of (1)

$=\left({C}_{0}\frac{{x}^{2}}{2}-{C}_{1}\frac{{x}^{3}}{3}+{C}_{2}\frac{{x}^{4}}{4}-{\dots +\frac{\left(-1{\right)}^{n}{C}_{n}}{n+1}{x}^{n+1}\right)]}_{0}^{1}$

$=\frac{{C}_{0}}{2}-\frac{{C}_{1}}{3}+\frac{{C}_{2}}{4}-\dots +\frac{\left(-1{\right)}^{n}{C}_{n}}{n+1}$

Thus,       $\frac{{C}_{0}}{2}-\frac{{C}_{1}}{3}+\frac{{C}_{2}}{4}-\dots +\left(-1{\right)}^{n}\frac{{C}_{n}}{n+1}$

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