Question

$\frac{1.2}{1!}+\frac{2.3}{2!}+\frac{3.4}{3!}+...\infty =xe$.

Find $x$

Answer 1

$T_n=\frac{n(n+1)}{n!}=\frac{n(n+1)}{(n-1)!}=\frac{(n-1)+2}{(n-1)!}$
$=\frac{1}{(n-2)!}+\frac{2}{(n-1)!}$
$\therefore$ Sum$=\sum _{n=2}^{\infty }\frac{1}{(n-2)!}+\sum _{n=1}^{\infty }\frac{2}{(n-1)!}$
$=\sum _{n=2}^{\infty }\frac{1}{(n-2)!}+2\sum _{n=1}^{\infty }\frac{1}{(n-1)!}=e+2e=3e$
Therefore $x=3$