First slide

Binomial theorem for positive integral Index

Question

If $C_{0}, C_{1}, C_{2}, \dots, C_{n}$ are the coefficients of the expansion of (1 + x) ⁿ , then the value of $\sum_{0}^{n} \frac{C_{k}}{k + 1}$ is

Moderate

A

0
B

$\frac{2^{n} - 1}{n}$
C

$\frac{2^{n + 1} - 1}{n + 1}$
D

none of these

Solution

Here, $t_{r + 1} = \frac{^{n} C_{r}}{r + 1} = \frac{1}{r + 1} \cdot^{n} C_{r}$
$= \frac{1}{n + 1} \cdot^{n + 1} C_{r + 1}$
Putting r = 0, 1, 2, ... n and adding, we get $\sum_{0}^{n} \frac{C_{k}}{k + 1}$
$\begin{aligned} = \frac{1}{n + 1} (^{n + 1} C_{1} +^{n + 1} C_{2} +^{n + 1} C_{3} + \dots +^{n + 1} C_{n + 1}) \\ = \frac{1}{n + 1} (2^{n + 1} -^{n + 1} C_{0}) = \frac{2^{n + 1} - 1}{n + 1} \end{aligned}$

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