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JEE Mains · Maths · STD 11 - basic of algoritham

माना \(\sum_{n=0}^{\infty} \frac{n^3((2 n) !)+(2 n-1)(n !)}{(n !)((2 n) !)}=a e+\frac{b}{e}+c\ \)है, जहाँ \(\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathbb{Z}\) तथा \(\mathrm{e}=\sum_{\mathrm{n}=0}^{\infty} \frac{1}{\mathrm{n} !}\) है तो \(\mathrm{a}^2-\mathrm{b}+\mathrm{c}\) बराबर है

  1. A \(25\)
  2. B \(24\)
  3. C \(23\)
  4. D \(26\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(26\)

Step-by-step Solution

Detailed explanation

\(\sum \limits_{n=0}^{\infty} \frac{n^3((2 n) !)+(2 n-1)(n !)}{(n !)((2 n) !)}\) \(=\sum \limits_{n=0}^{\infty} \frac{1}{(n-3) !}+\sum \limits_{n=0}^{\infty} \frac{3}{(n-2) !}\)…
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