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) .

Both are equal to the number of ways of selecting three numbers from $1, \ldots, n+2$ : the first one by definition, the second by choosing the middle number, say $i+1$, and then by choosing one of the $i$ available "to the left" to be the smaller one, and one of the $(n+2)-(i+1)=n+1-i$ available "to the right" to be the bigger one.
