<s>[INST] Prove using mathematical induction: $1 \cdot 2 \cdot 3 + \ldots + n \cdot (n+1) \cdot (n+2) =  \frac{n(n+1)(n+2)(n+3)}{4}$ So I have the easy stuff done. However, I'm not sure how to go about doing the inductive step with such an awkward proof. :/
Statement: 
$$ 1 \cdot 2 \cdot 3 + … + n \cdot (n+1) \cdot (n+ 2) = \frac{n \cdot (n + 1) \cdot (n + 2) \cdot (n +3)}{4} .$$
Base step: $n = 1$: 
$$ 0  \cdot (0 + 1) \cdot (0 + 2) + 1 \cdot (1 + 1) \cdot (1 +2) = \frac{ 1 \cdot (1+ 1) \cdot (1 + 2) \cdot (1 + 3) }{4} .$$
$$ 0 \cdot (0 + 1) \cdot (0 + 2) +1 \cdot (1 + 1) \cdot (1 +2) = 0 + 6 = 6$$
$$ \frac{ 1 \cdot (1+ 1) \cdot (1 + 2) \cdot (1 + 3) }{4} = \frac{24}{ 4} = 6 .$$
$$6 = 6.$$
Inductive step:
$n = n + 1$
 [/INST] For the record, the "correct" way to prove this identity is combinatorially. Written differently (after dividing both sides by $6$), it reads
$$ \sum_{k=3}^{n+2} \binom{k}{3} = \binom{n+3}{4}. $$
This is because each $4$-subset of $[n+3]$ is composed of its maximum $k+1$ and a $3$-subset of $[k]$.
</s>