How to prove ${n+2 \choose 3}=1\cdot n + 2 \cdot (n - 1) + \ldots + n \cdot 1$? I saw this problem as an exercise in Combinatorial Identities :-

Prove that $${n+2 \choose 3}=1\cdot n + 2 \cdot (n - 1) + \ldots + n \cdot 1\,.$$

After giving some time to this, I think that it is quite similar to the identity :-
${n \choose k}  = {n - 1 \choose k - 1} + {n - 1 \choose k}$ 
But I don't know how to prove this algebraically , anyone please help me with this.
(Note that I am still not sure whether we can use that identity or not , I can also guess we can use Vandermonde's Identity here) .

I suggest proving it combinatorially. $\binom{n+2}3$ is the number of $3$-element subsets of the set $[n+2]=\{1,2,\ldots,n+2\}$. We can classify those sets by their middle elements: let $\mathscr{A}_k$ be the family of all $3$-element subsets of $[n+2]$ of the form $\{j,k,\ell\}$, where $j<k<\ell$; clearly
$$\binom{n+2}3=\sum_k|\mathscr{A}_k|\;.$$
Now prove that $|\mathscr{A}_k|=(k-1)(n+2-k)$ and determine the range of possible values of $k$ to complete the proof.
