ExamBro
ExamBro
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 :

  1. A \(\frac{11 e }{2}+\frac{7}{2 e }\)
  2. B \(\frac{13 e }{4}+\frac{5}{4 e }-4\)
  3. C \(\frac{11 e }{2}+\frac{7}{2 e }-4\)
  4. D \(\frac{13 e }{4}+\frac{5}{4 e }\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{13 e }{4}+\frac{5}{4 e }-4\)

Step-by-step Solution

Detailed explanation

\(\sum \limits_{ n =1}^{\infty} \frac{2 n ^2+3 n +4}{(2 n ) !}\) \(\frac{1}{2} \sum \limits_{ n =1}^{\infty} \frac{2 n (2 n -1)+8 n +8}{(2 n ) !}\)…