ExamBro
ExamBro
JEE Mains · Maths · STD 11 - basic of algoritham

Let \(\quad \sum \limits_{n=0}^{\infty} \frac{n^3((2 n) !)+(2 n-1)(n !)}{(n !)((2 n) !)}=a e+\frac{b}{e}+c\), where \(a, b, c \in Z\) and \(e=\sum \limits_{n=0}^{\infty} \frac{1}{n!}\) Then \(a^2-b+c\) is equal to \(................\).

  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

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