NIMCET 2024 — Mathematics PYQ

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Question

NIMCET 2024 — Mathematics PYQ

NIMCET | Mathematics | 2024

The value of series

$\frac{2}{3 !} + \frac{4}{5 !} + \frac{6}{7 !} + \dots$ is

Choose the correct answer:

Explanation

$T_r = \frac{2r}{(2r+1)!} = \frac{(2r+1) - 1}{(2r+1)!}$

$T_r = \frac{1}{(2r)!} - \frac{1}{(2r+1)!}$

$S = \sum_{r=1}^{\infty} \left( \frac{1}{(2r)!} - \frac{1}{(2r+1)!} \right)$

$S = \left( \frac{1}{2!} - \frac{1}{3!} \right) + \left( \frac{1}{4!} - \frac{1}{5!} \right) + \left( \frac{1}{6!} - \frac{1}{7!} \right) + \dots$

We know:

$e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots$

$e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dots$

Subtracting the two series:

$e - e^{-1} = 2 \left( \frac{1}{1!} + \frac{1}{3!} + \frac{1}{5!} + \dots \right)$

Adding the two series:

$e + e^{-1} = 2 \left( 1 + \frac{1}{2!} + \frac{1}{4!} + \dots \right)$

Given the option $e - 1$ , let us re-examine the series expansion:

If the series is $\sum_{n=1}^{\infty} \frac{n+1}{(n+1)!} = \sum \frac{1}{n!}$

$S = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$

$S = e - 1$

Correct Option: (C)