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

A
$\frac{1}{n + 1}$
B
$\frac{1}{n + 2}$
C
$\frac{1}{n (n + 1)}$
D
$\frac{1}{(n + 1) (n + 2)}$

Solution:

We have

$\begin{array}{l} \Rightarrow (1 - x)^{n} = C_{0} - C_{1} x + C_{2} x - C_{3} x + \dots + (- 1)^{n} C_{n} x^{n} \\ \Rightarrow \int_{0}^{1} x (1 - x)^{n} d x = \int_{0}^{1} [C_{0} x - C_{1} x^{2} + C_{2} x^{3} - C_{3} x^{4} + \dots \\ + (- 1)^{n} C_{n} x^{n + 1}] d x \underset{}{\to (1)} \end{array}$

L.H.S. of $(1)$

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

R.H.S of (1)

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

$= \frac{C_{0}}{2} - \frac{C_{1}}{3} + \frac{C_{2}}{4} - \dots + \frac{(- 1)^{n} C_{n}}{n + 1}$

Thus, $\frac{C_{0}}{2} - \frac{C_{1}}{3} + \frac{C_{2}}{4} - \dots + (- 1)^{n} \frac{C_{n}}{n + 1}$

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

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

Solution:

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