Question

$\frac{2}{2!}+\frac{2+4}{3!}+\frac{2+4+6}{4!}+\dots ..=$

• A. $e-2$
• B. $e-1$
• C. $e$
• D. $1/e$

Answer 1

The correct option is C $e$

$=\sum _{n=1}^{\infty }\frac{2+4+6---2\left(n\right)}{n!}$
$=2\sum _{n=1}^{\infty }\frac{(1+2+3---(n\left)\right)}{\left(n\right)!}$
$=2\sum _{n=1}^{\infty }\frac{n(n-1)}{2n!}$
$=\sum _{n=1}^{\infty }\frac{1}{(n-1)!}\dots \dots (e^x=\sum _{n=0}^{\infty }\frac{x^n}{n!})$
$=e$