Q. The sum $\sum_{n=1}^{\infty}\frac{2n^2+3n+4}{(2n)!}$ is equal to :
A) $\frac{11e}{2}+\frac{7}{2e}$
B) $\frac{13e}{4}+\frac{5}{4e}-4$
C) $\frac{11e}{2}+\frac{7}{2e}-4$
D) $\frac{13e}{4}+\frac{5}{4e}$
Explanation
The correct answer is Option B: $\frac{13e}{4}+\frac{5}{4e}-4$.
A detailed explanation for this question is being prepared.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.