Question

$\frac{2}{1!}+\frac{2+4}{2!}+\frac{2+4+6}{3!}+\frac{2+4+6+8}{4!}+...\infty =ae$. Find $a$

Answer 1

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

$\therefore$ 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 $a=3$