Question

The sum $\displaystyle\sum_{n=1}^{\infty} \frac{2 n^2+3 n+4}{(2 n) !}$ is equal to:

• A. $\frac{11 e }{2}+\frac{7}{2 e }-4$
• B. $\frac{11 e }{2}+\frac{7}{2 e }$
• C. $\frac{13 e }{4}+\frac{5}{4 e }-4$
• D. $\frac{13 e }{4}+\frac{5}{4 e }$

Answer 1

The Correct Option is C

\(\displaystyle\sum_{n=1}^{∞}\frac {2n2+3n+4}{(2n)!}\)

\(\frac 12 \displaystyle\sum_{n=1}^{∞}\frac {2n(2n−1)+8n+8}{(2n)!}\)

\(\frac 12\displaystyle\sum_{n=1}^{∞}\frac {1}{(2n−2)!}+2\displaystyle\sum_{n=1}^{∞}\frac {1}{(2n−1)!}+4\displaystyle\sum_{n=1}^{∞}\frac {1}{(2n)!}\)

\(e=1+1+\frac {1}{2!}+\frac {1}{3!}+\frac {1}{4!}+….\)

\(e^{-1}=1-1+\frac {1}{2!}-\frac {1}{3!}+\frac {1}{4!}+….\)

\((e+\frac 1e)=2(1+\frac {1}{2!}+\frac {1}{4!}+…….)\)

\(e−\frac 1e=(1+\frac {1}{3!}+\frac {1}{5!}+…..)\)
Now,
\(\frac 12(\displaystyle\sum_{n=1}^{∞}\frac {1}{(2n−2)!})+2\displaystyle\sum_{n=1}^{∞}\frac {1}{(2n−1)!}+4\displaystyle\sum_{n=1}{∞}\frac {1}{(2n)!}\)

\(=\frac 12[\frac {e+\frac 1e}{2}]+2[\frac {e−\frac 1e}{2}]+4(\frac {e+\frac 1e−2}{2})\)

\(=\frac {(e+\frac 1e)}{4}+e−\frac 1e+2e+\frac 2e−4\)

\(=\frac {13e}{4}+\frac {5}{4e}−4\)

Therefore, the correct option is (C): \(\frac {13e}{4}+\frac {5}{4e}−4\).