In need of a combinatorial proof of $1*2 + 2*3 + 3*4 + ... + n(n+1) = 2*{n+2\choose 3}$

Question

$$1*2 + 2*3 + 3*4 + ... + n(n+1) = 2*{n+2\choose 3}$$ for natural numbers $n$.

For my discrete maths class, we are learning combinatorial proofs, and my prof handed this problem as practice. I'm aware of the process of a combinatorial proof (counting the same thing in two different ways), but I have no idea how to start this problem or how to relate the LHS to a counting problem.

All I know is the RHS is twice the number of ways to make $3$ element subsets of a set of $n+2$ elements. A possible hint I found from a related problem in my book is to consider a set $$S = \{0, 1, 2, 3, ... , n, n+1\}$$ which has $n+2$ elements.

Answer 1

As in @YiFan's comment, it's easier if we divide everything by $2$. Suppose you want to pick a $3$-element subset of $\{1,2,\dots,n+2\}$. Two ways to do this are:

• Pick all three elements simultaneously. There are $\binom{n+2}{3}$ ways to do this, by definition.
• Pick the largest element first. If the largest element is $k$, then there are $\binom{k-1}{2}$ ways to pick the remaining elements. Then sum over all possible values of $k$.