First slide

Binomial theorem for positive integral Index

Question

The value of $\frac{^{n} C_{1}}{2} + \frac{^{n} C_{3}}{4} + \frac{^{n} C_{5}}{6} + \dots$ is

Moderate

A

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

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

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

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

Solution

The rth term of the given expression is
$\begin{aligned} T_{r} = \frac{^{n} C_{2 r - 1}}{2 r} \\ Since \frac{1}{r + 1} \cdot^{n} C_{r} = \frac{1}{n + 1} \cdot^{n + 1} C_{r + 1} \\ ∴ T_{r} = \frac{^{n} C_{2 r - 1}}{2 r} = \frac{1}{n + 1} \cdot^{n + 1} C_{2 r} \\ ∴ \frac{^{n} C_{1}}{2} + \frac{^{n} C_{3}}{4} + \frac{^{n} C_{5}}{6} + \dots \end{aligned}$
$= \frac{1}{n + 1} (^{n + 1} C_{2} +^{n + 1} C_{4} + \dots)$
$\begin{aligned} = \frac{1}{n + 1} (2^{n + 1 - 1} -^{n + 1} C_{0}) \\ = \frac{2^{n} - 1}{n + 1} \end{aligned}$

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