Find the sum of the series
`(1^(2)+1)1!+(2^(2)+1)2!+(3^(2)+1)3!+ . .+(n^(2)+1)n!`.

Explanation

Let `S_(n)=(1^(2)+1)1!+(2^(2)+1)2!+(3^(2)+1)3!+ . .+(n^(2)+1)n!`
`therefore` nth term `T_(n)=(n^(2)+1)n!`
`={(n+1)(n+2)-3(n+1)+2}n!`
`T_(n)=(n+2)!-3(n+1)!+2n!`
Putting `n=1,2,3,4, . .,n`
then `T_(1)=3!-3*2!+2*1!`
`T_(2)=4!-3*3!+2*2!`
`T_(3)+%!-3*4!+2*3!`
`T_(4)=6!-3*5!+2*4!`
. . . . . . .
. . . . . . .
`T_(n-1)=(n+1)!-3n!+2(n-1)!`
`T_(n)=(n+2)!-3(n+1)!+2n!`
`therefore S_(n)=T_(1)+T_(2)+T_(3)+ . .. +T_(n)`
`=(n+2)!-2(n+1)!` [the rest cancel out]
`=(n+2)(n+1)!-2(n+1)!`
`=(n+1)!(N=2-2)`
`=n(n+1)!`