Question

The sum of the series $1+\frac{1^2+2^2}{2!}+\frac{1^2+2^2+3^3}{3!}+\frac{1^2+2^2+3^2+4^2}{4!}+.....$ is

• A. 3e
• B. $\frac{17}{6}e$
• C. $\frac{13}{6}e$
• D. $\frac{19}{6}e$

Answer 1

Answer: (B)

$\frac{17}{6}e$

$T_n=\frac{1^2+2^2+3^2+......+n^2}{n!}$
$\frac{\sum n^2}{n!}=\frac{n(n+1)(2n+1)}{6n!}$
$=\frac{1}{6}\left(\frac{2n^3+3n^2+n}{n!}\right)$
$=\frac{1}{6}(\frac{2n^3}{n!}+\frac{3n^2}{n!}+\frac{n}{n!})$
$\therefore$ sum of the series
$=\frac{1}{6}(2\sum _{n=1}^{\infty }\frac{n^3}{n!}+3\sum _{n=1}^{\infty }\frac{n^2}{n!}+\sum _{n=1}^{\infty }\frac{n}{n!})$
$=\frac{1}{6}(2\times 5e+3\times 2e+e)$
$=\frac{1}{6}(10e+6e+e)=\frac{17}{6}e$

Hence, option B is correct.