Question

The value of $1\cdot 1!+2\cdot 2!+3\cdot 3!+...+n\cdot n!$ is

• A. $(n+1)!$
• B. $(n+1)!+1$
• C. $(n+1)!-1$
• D. None of these

Answer 1

The correct option is C $(n+1)!-1$

We have $1\cdot 1!+2\cdot 2!+3\cdot 3!+...+n\cdot n!$
$={\sum }_{r=1}^{n}r(r!)={\sum }_{r=1}^n\left[\right(r+1)r!-r!]$
$={\sum }_{r=1}^n(r+1)!-r!=(2!-1!)+(3!-2!)+....\left[\right(n+1)!-n!]$
$={\sum }_{r=1}^n\left[\right(r+1)!-r!]=(2!-1!)+(3!-2!)+....\left[\right(n+1)!-n!]$
$=(n+1)!-1!=(n+1)!-1$