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WBJEE · Maths · Permutation Combination

The value of \(\frac{2}{3 !}+\frac{4}{5 !}+\frac{6}{7 !}+\ldots \ldots\). is

  1. A \(\mathrm{e}^{\frac{1}{2}}\)
  2. B \(\mathrm{e}^{-1}\)
  3. C \(\mathrm{e}\)
  4. D \(\mathrm{e}^{-\frac{1}{3}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\mathrm{e}^{-1}\)

Step-by-step Solution

Detailed explanation

\begin{gathered} \text { Hints: } t_n=\frac{2 n}{(2 n+1) !}=\frac{2 n+1}{(2 n+1) !}-\frac{1}{(2 n+1) !}=\frac{1}{(2 n) !}-\frac{1}{(2 n+1) !} \\ \sum_{n=1}^{\infty} t_n=\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\frac{1}{5 !}+\ldots \ldots \infty=\mathrm{e}^{-1} \end{gathered}