Aloha :)
Zu zeigen: \(S_n:=\sum\limits_{k=1}^nk(k+1)(k+2)=\frac{n(n+1)(n+2)(n+3)}{4}\)
Verankerung bei \(n=1\):$$\sum\limits_{k=1}^1k(k+1)(k+2)=1\cdot2\cdot3=6=\frac{24}{4}=\frac{1\cdot2\cdot3\cdot4}{4}\quad\checkmark$$Induktionsschritt \(n\to n+1\):
$$S_{n+1}=\sum\limits_{k=1}^{n+1}k(k+1)(k+2)$$$$\phantom{S_{n+1}}=(n+1)((n+1)+1)((n+1)+2)+\sum\limits_{k=1}^nk(k+1)(k+2)$$$$\phantom{S_{n+1}}\stackrel{I.V.}{=}(n+1)(n+2)(n+3)+\frac{n(n+1)(n+2)(n+3)}{4}$$$$\phantom{S_{n+1}}=\frac{4(n+1)(n+2)(n+3)}{4}+\frac{n(n+1)(n+2)(n+3)}{4}$$$$\phantom{S_{n+1}}=\frac{4\overbrace{(n+1)(n+2)(n+3)}^{ausklammern}+n\,\overbrace{(n+1)(n+2)(n+3)}^{ausklammern}}{4}$$$$\phantom{S_{n+1}}=\frac{(4+n)\overbrace{(n+1)(n+2)(n+3)}^{ausgeklammert}}{4}$$$$\phantom{S_{n+1}}=\frac{(n+1)(n+2)(n+3)(n+4)}{4}\quad\checkmark$$
