Question

Cum se calculează suma de la exercițiul 11?

Answer 1

$\sum_{k=1}^{n}(k+1)^2k!=\sum_{k=1}^{n}(k+1)k!(k+1)=\sum_{k=1}^{n}(k+1)!(k+1)=\\\\=\sum_{k=1}^{n}(k+1)!(k+2-1)=\sum_{k=1}^{n}(k+1)!(k+2)-\sum_{k=1}^{n}(k+1)!=\\\\=\sum_{k=1}^{n}(k+2)!-\sum_{k=1}^{n}(k+1)!=3!-2!+4!-3!+\ldots+(n+1)!-\\\\-n!+(n+2)!-(n+1)!=(n+2)!-2.$

Green eyes.