How do you simplify $(n - 1)! (n^{2} + n)$ $(n-1)!(n^2+n)$ ?

Question

6938 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-18T16:57:35

$(n - 1)! (n^{2} + n) = (n + 1)!$ $(n-1)!(n^2+n)=(n+1)!$

$(n - 1)! (n^{2} + n)$ $(n-1)!(n^2+n)$

= $(n - 1)! \times n \times (n + 1)$ $(n-1)!×n×(n+1)$

= $(n + 1) \times n \times (n - 1)!$ $(n+1)×n×(n-1)!$

= $(n+1)×n×(n-1)×(n-2)×(n-3)×...×3×2×1$

= $(n+1)!$

How do you simplify (n-1)!(n^2+n)(n−1)!(n2+n)(n-1)!(n^2+n)?