JEE Mains · Maths · STD 11 - basic of algoritham
The sum \(\sum \limits_{n=1}^{\infty} \frac{2 n^2+3 n+4}{(2 n) !}\) is equal to :
- A \(\frac{11 e }{2}+\frac{7}{2 e }\)
- B \(\frac{13 e }{4}+\frac{5}{4 e }-4\)
- C \(\frac{11 e }{2}+\frac{7}{2 e }-4\)
- D \(\frac{13 e }{4}+\frac{5}{4 e }\)
Answer & Solution
Correct Answer
(B) \(\frac{13 e }{4}+\frac{5}{4 e }-4\)
Step-by-step Solution
Detailed explanation
\(\sum \limits_{ n =1}^{\infty} \frac{2 n ^2+3 n +4}{(2 n ) !}\) \(\frac{1}{2} \sum \limits_{ n =1}^{\infty} \frac{2 n (2 n -1)+8 n +8}{(2 n ) !}\)…
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