Question

Solve : $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+.......+\frac{2012}{2013!}$

• A. $1+\frac{1}{2012!}$
• B. $\frac{2013}{2!3!...2012!}$
• C. $1-\frac{1}{2013!}$
• D. $\frac{2014}{2013!}$

Answer 1

The correct option is C $1-\frac{1}{2013!}$

Given $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...\frac{2014}{2013!}$
$=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...\frac{2013-1}{2013!}$
$=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{2012!}-\frac{1}{2013!}$

Hence after simplifying, we get

$=1-\frac{1}{2013!}$