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 \(................\).
- A \(25\)
- B \(24\)
- C \(23\)
- D \(26\)
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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