Question

$\sum _{r=1}^{99}r!(r^2+r+1)=$

• A. $102!-100!$
• B. $100(100!)-1$
• C. $99(100!)-1$
• D. $100(99!)-1$

Answer 1

The correct option is B $100(100!)-1$

$r!(r^2+r+1)=r!\left[\right(r+1{)}^2-r]$
$=r!(r+1)(r+1)-r!r$
$=(r+1)!(r+1)-r!r$
Which is of form $V_{r+1}-V_r$
$\therefore \sum _{r=1}^{99}r!(r^2+r+1)=V_{100}-V_1$
$=(100!)\cdot 100-1$