Sum of the series `underset(r=1)overset(n)(sum)(r^(2)+1)r!` is

Options

A. `(n+1)!`
B. `(n+2)!-1`
C. `n(n+1)!`
D. none of these

Explanation

We have,
`sum_(r=1)^(n)(r^(2)+1)r!`
`sum_(r=1)^(n){(r+2)(r+1)-3(r+1)+2}r!`
`sum_(r=1)^(n){(r+2)!-(r+1)!}-2sum_(r=1)^(n){(r+1)!-r!}`
`=(n+2)!-2!-2{(n +1)!-1}=n(n+1)!`