Question

$1+\frac{1^2+2^2}{2!}+\frac{1^2+2^2+3^2}{3!}+\frac{1^2+2^2+3^2+4^2}{4!}+...\infty =\frac{b}{6}e$ . Find $b$

Answer 1

Let $T_n$ be the nth term of the given series then
$T_n=\frac{1^2+2^2+3^2+4^2+....+n^2}{n!}$

$=\frac{\sum n^2}{n!}$

$=\frac{n(n+1)(2n+1)}{6n(n-1)!}$ .... (sum of squares of the first n natural numbers)

$=\frac{(n+1)(2n+1)}{6(n-1)!}$

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

Dividing $2n^2+3n+1$ by $n-1$

$\therefore$ $2n^2+3n+1=(n-1)(2n+5)+6$

$\therefore$ $T_n=\frac{(n-1)(2n+5)+6}{6(n-1)!}$

$=\frac{(n-1)(2n+5)}{6(n-1)!}+\frac{1}{(n-1)!}$

$=\frac{(n-1)(2n+5)}{6(n-1)(n-2)!}+\frac{1}{(n-1)!}$

$=\frac{(2n+5)}{6(n-2)!}+\frac{1}{(n-1)!}$

Dividing $2n+5$ by $n-2$

$\therefore$ $2n+5=2(n-2)+9$

$\therefore$ $T_n=\frac{2(n-2)+9}{6(n-2)!}+\frac{1}{(n-1)!}$

$=\frac{2(n-2)}{6(n-2)!}+\frac{9}{6(n-2)!}+\frac{1}{(n-1)!}$

$=\frac{2}{6(n-3)!}+\frac{9}{6(n-2)!}+\frac{1}{(n-1)!}$

Hence sum of infinite terms of given series

$S_{\infty }=\sum _{n=1}^{\infty }{T}_n$

$=\sum _{n=1}^{\infty }(\frac{2}{6(n-3)!}+\frac{9}{6(n-2)!}+\frac{1}{(n-1)!})$

$=\frac{2}{6}\sum _{n=1}^{\infty }\frac{1}{(n-3)!}+\frac{9}{6}\sum _{n=1}^{\infty }\frac{1}{(n-2)!}+\sum _{n=1}^{\infty }\frac{1}{(n-1)!}$

$=\frac{2}{6}(\frac{1}{(-2)!}+\frac{1}{(-1)!}+\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...)+\frac{9}{6}(\frac{1}{(-1)!}+\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...)+(\frac{1}{0!}+\frac{1}{2!}+\frac{1}{3!}+...)$

$=\frac{2}{6}(0+0+e)+\frac{9}{6}(0+e)+\left(e\right)$ . ..... $e^x=1+x+\frac{x^2}{2!}+\frac{x3}{3!}+...$

$=\frac{17}{6}e$

$\Rightarrow b=17$