Prove that $\sum_{r=1}^nr(r+1)=\frac{n(n+1)(n+2)}{3}$ using induction

Question

$$\sum_{r=1}^nr(r+1)=\frac{n(n+1)(n+2)}{3}$$

could you help me with how exactly I work this out?

Answer 1

Hint:-

$\dfrac{(k+1)(k+2)(k+3)}{3}-\dfrac{k(k+1)(k+2)}{3}\\= (k+1)(k+2)\\=\displaystyle\sum_{r=1}^{k+1}r(r+1)-\displaystyle\sum_{r=1}^{k}r(r+1)$

Answer 2

Hint: $$\sum_{r=1}^kr(r+1)=\frac{k(k+1)(k+2)}{3}$$

$$\begin{align} \sum_{r=1}^{k+1}r(r+1) &=(k+1)(k+2)+\sum_{r=1}^kr(r+1)\\ &=(k+1)(k+2)+\frac{k(k+1)(k+2)}{3}\end{align}$$

Now simplify

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otherwise use $$\sum_{r=1}^nr(r+1)=\sum_{r=1}^n(r^2+r)=\sum_{r=1}^nr^2+\sum_{r=1}^nr$$