Question:

$ \frac{1}{3!}+\frac{2}{5!}+\frac{3}{7!}+... $ is equal to

Updated On: Jun 23, 2024
  • $ \frac{{{e}^{-1}}}{2} $
  • $ e $
  • $ \frac{e}{4} $
  • $ \frac{e}{6} $
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The Correct Option is A

Solution and Explanation

Let $ S=\frac{1}{3!}+\frac{2}{5!}+\frac{3}{7!}+..... $
$ \therefore $ $ {{T}_{n}}=\frac{n}{(2n+1)!} $
$ =\frac{1}{2}\left[ \frac{2n+1-1}{(2n+1)!} \right]=\frac{1}{2}\left[ \frac{1}{(2n)!}-\frac{1}{(2n+1)!} \right] $
$ \therefore $ $ {{T}_{1}}=\frac{1}{2}\left( \frac{1}{2!}-\frac{1}{3!} \right) $
$ {{T}_{2}}=\frac{1}{2}\left( \frac{1}{4!}-\frac{1}{5!} \right) $
$ \therefore $ $ S={{T}_{1}}+{{T}_{2}}+.... $
$ =\frac{1}{2}\left[ \frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+.....\infty +1-1 \right] $
$ =\frac{{{e}^{-1}}}{2} $
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Concepts Used:

Permutations

A permutation is an arrangement of multiple objects in a particular order taken a few or all at a time. The formula for permutation is as follows:

\(^nP_r = \frac{n!}{(n-r)!}\)

 nPr = permutation

 n = total number of objects

 r = number of objects selected

Types of Permutation

  • Permutation of n different things where repeating is not allowed
  • Permutation of n different things where repeating is allowed
  • Permutation of similar kinds or duplicate objects