Question

2! pe 1! + 3! pe 2! +...+(n+1)! pe n!

Answer 1

$\dfrac{2!}{1!}+\dfrac{3!}{2!}+... + \dfrac{(n+1)!}{n!}= \\ \\ =\dfrac{1!\cdot 2}{1!}+\dfrac{2!\cdot 3}{2!}+... + \dfrac{n!\cdot (n+1)}{n!}=\\ \\ =2+3+...+(n+1) =\\ \\= 1+2+3+...+(n+1) -1 = \\ \\ = \dfrac{(n+1)\Big[(n+1)+1\Big]}{2} -1 \\ \\= \dfrac{(n+1)(n+2)}{2} -1$