First slide

Binomial theorem for positive integral Index

Question

$^{n} C_{0} - {\frac{1}{2}}^{n} C_{1} + {\frac{1}{3}}^{n} C_{2} - \dots + (- 1)^{n} \frac{^{n} C_{n}}{n + 1}$ is equal to

Moderate

A

n
B

$\frac{1}{n}$
C

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

$\frac{1}{n - 1}$

Solution

We know, $(1 - x)^{n} =^{n} C_{0} -^{n} C_{1} x +^{n} C_{2} x^{2} \dots {(- 1)^{n}}^{n} C_{n} x^{n}$ On integrating limit 0 to 1, we get
$\int_{0}^{1} (1 - x)^{n} dx = \int_{0}^{1} (^{n} C_{0} -^{n} C_{1} x +^{n} C_{2} x^{2} \dots (- 1)^{nn} C_{n} x^{n}) dx$ $\to {[\frac{- (1 - x)^{n + 1}}{n + 1}]}_{0}^{1}$
$\begin{aligned} = {[^{n} C_{0} x - \frac{^{n} C_{1} x^{2}}{2} + \frac{^{n} C_{2} x^{3}}{3} \dots \frac{(- 1)^{nn} C_{n} x^{n - 1}}{n + 1}]}_{0}^{1} \\ \Rightarrow 0 + \frac{1}{n + 1} =^{n} C_{0} - \frac{^{n} C_{1}}{2} + \frac{^{n} C_{2}}{3} - \dots \frac{(- 1)^{n n} C_{n}}{n + 1} \end{aligned}$

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