Question

The value of the infinite series $\frac{1^2+2^2}{3!}+\frac{1^2+2^2+3^2}{4!}+\frac{1^2+2^2+3^2+4^2}{5!}+......$ is

• A. $e$
• B. $5e$
• C. $\frac{5e}{6}-\frac{1}{2}$
• D. $\frac{5e}{6}$

Answer 1

The correct option is C $\frac{5e}{6}-\frac{1}{2}$

Given infinite series is $\frac{1^2+2^2}{3!}+\frac{1^2+2^2+3^2}{4!}+\frac{1^2+2^2+3^2+4^2}{5!}+......$
$n^{th}$ term $t_n=\frac{1^2+2^2+3^2+4^2+.....+{(r+1)}^2}{(r+2)!}$

Now

$S_n=\sum t_n={\sum }_{r=1}^n\frac{1^2+2^2+3^2+4^2+.....+{(r+1)}^2}{(r+2)!}$

$={\sum }_{r=1}^n\frac{(r+1)(r+2)(2n+3)}{6(r+2)!}[\because \sum n^2=\frac{n(n+1)(2n+1)}{6}]$

$={\sum }_{r=1}^n\frac{(2r+3)}{6r!}=\frac{1}{6}{\sum }_{r=1}^n\{\frac{2}{(r-1)!}+\frac{3}{r!}\}$

$=\frac{1}{6}\{(\frac{2}{1}+\frac{3}{1!})+(\frac{2}{1!}+\frac{3}{2!})+(\frac{2}{2!}+\frac{3}{3!})+...\}$

$=\frac{1}{6}\{(\frac{2}{1!}+\frac{2}{1!}+\frac{2}{2!}+..)+(\frac{3}{1!}+\frac{3}{2!}+\frac{3}{3!})\}$

$=\frac{1}{6}\{2(1+\frac{1}{1!}+\frac{1}{2!}+..)+3(1+\frac{1}{1!}+\frac{1}{2!}+.)\}$

$=\frac{1}{6}2e+3(e-1)$

$=\frac{1}{6}(2e+3e-3)$

$=\frac{1}{6}(5e-3)=\frac{5e}{6}-\frac{1}{2}$