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

Question

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?

Answer 1

Hint

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

Answer 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 $$2\binom{n+2}{3}+(n+1)(n+2)=\frac{2n(n+1)(n+2)}{6}+(n+1)(n+2)=$$ $$(n+1)(n+2)(n/3+1)=2\frac{(n+3)(n+2)(n+1)}{6}=2\binom{n+3}{3}$$

Answer 3

I'll try and explain it by using another formula. Suppose we have $ \sum_{i=1}^n i $. I claim that this sum is equal to $n(n+1)/2$. How did I manage to come up with this ? Well, I just looked at the numbers and their sum and tried to find a pattern. Now to justify that this is indeed correct, I will use induction. So first you claim and then you prove. Try to come up with such a formula, break down the summation into two parts and then look for a pattern.

Answer 4

Note that $$1\times 2=\frac{1\times2\times3}{3},$$ and $$1\times 2 + 2\times 3 = \frac{2\times 3 \times 4}{3}.$$ And in general what you want to prove is $$\sum_{k=1}^{n}k(k+1)=\frac{n(n+1)(n+2)}{3}.$$