Induction problem: a formula for $\sum_{i=1}^n i(i+1)$

Find a formula for $\sum_{i=1}^n i(i+1)$ and prove that it holds for all $n \geq$ 1.

For this induction problem I chose $i = n +1$ so we have $(n+1)(n+1 +1) = (n+1)(n+2)$. Is that what we suppose to get for this problem?

Solution 1:

Hint

Use the binomial identity $$ {2 \choose 2} + {3 \choose 2} + {4 \choose 2} + \cdots + {k \choose 2} = {k+1 \choose 3}. $$

Solution 2:

$$\sum_{i=1}^{n}i(i+1)=\sum_{i=1}^{n}(i^2+i)=\sum_{i=1}^{n}i^2+\sum_{i=1}^{n}i=$$ $$=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}=\frac{n(n+1)}{2}\frac{2n+4}{3}=$$ $$=\frac{2n(n+1)(n+2)}{6}=2\binom{n+2}{3}$$ hint for induction proof $$2\binom{n+2}{3}+(n+1)(n+2)=\frac{2n(n+1)(n+2)}{6}+(n+1)(n+2)=$$ $$(n+1)(n+2)(n/3+1)=2\frac{(n+3)(n+2)(n+1)}{6}=2\binom{n+3}{3}$$