Proving $1 \cdot 2 + 2 \cdot 3 + \cdots + n(n + 1) = 2\binom{n + 2}{3}$by math induction?

Question

I am working on a problem, but I don't know whether or not to use math induction on it. Here's the problem:

Prove that for all integers $n \geq 1$,

$$1 \cdot 2 + 2 \cdot 3 + \cdots + n(n + 1) = 2\binom{n + 2}{3}$$

If I used math induction, I'd first pick out a basis which would be $1$. Then I'd substitute $n$ for $1$. Which would be:

$$1(1 + 1) = 2\binom{1+2}{3}$$

the left hand side is equal to the right hand side, thus we have verified this part of the proof.

Next we look at $n + 1$: $$n + 1(n + 2) = 2\binom{n + 3}{3}$$

How do we prove the last part?

Answer 1

Check carefully and algebraically each step:

$$\overbrace{1\cdot2+2\cdot3+\ldots+n(n+1)}^{=2\binom{n+2}3}+(n+1)(n+2)=2\frac{(n+2)!}{3!(n-1)!}+(n+1)(n+2)=$$

$$=\frac{2(n+2)!n+6(n+2)!}{6n!}=\frac{(n+2)!}{3n!}\left(n+3\right)=2\frac{(n+3)!}{3!n!}=2\binom{n+3}3$$

Answer 2

Hint : After finding that the equation is true for n=1, you then assume that the given equation is true for some positive n greater than or equal to 1.

and then using this assumption you prove that the same equation holds for n+1.