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

Answer: (C)

$\frac{13 e }{4}+\frac{5}{4 e }-4$

Answer 2

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

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

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

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

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

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

$e−\frac 1e=(1+\frac {1}{3!}+\frac {1}{5!}+…..)$
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): $\frac {13e}{4}+\frac {5}{4e}−4$.