First slide

Binomial theorem for positive integral Index

Question

If $C_{0}, C_{1}, C_{2}, \dots, C_{n}$ denote the binomial coefficients in the expansion of $(1 + x)^{n}$ , then
$C_{0} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + \dots + \frac{C_{n}}{n + 1} or, \sum_{r = 0}^{n} \frac{C_{r}}{r + 1}$

Moderate

A

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

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

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

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

Solution

We have
$\begin{array}{l} C_{0} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + \dots + \frac{C_{n}}{n + 1} \\ = \sum_{r = 0}^{n} \frac{C_{r}}{r + 1} \\ = \sum_{r = 0}^{n} \frac{1}{r + 1} \cdot^{n} C_{r} \\ = \sum_{r = 0}^{n} \frac{1}{n + 1} \cdot \frac{n + 1}{r + 1} \cdot^{n} C_{r} \\ = \frac{1}{n + 1} \sum_{r = 0}^{n} \frac{n + 1}{r + 1} \cdot^{n} C_{r} \end{array}$
$\begin{array}{l} = \frac{1}{n + 1} \sum_{r = 0}^{n}^{n + 1} C_{r + 1} [∵^{n + 1} C_{r + 1} = \frac{n + 1}{r + 1} \cdot^{n} C_{r}] \\ = \frac{1}{n + 1} \{^{n + 1} C_{1} +^{n + 1} C_{2} +^{n + 1} C_{3} + \dots +^{n + 1} C_{n + 1}\} \\ = \frac{1}{n + 1} \{(^{n + 1} C_{0} +^{n + 1} C_{1} + \dots +^{n + 1} C_{n + 1}) - (^{n + 1} C_{0})\} \\ = \frac{1}{n + 1} (2^{n + 1} - 1) \end{array}$

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