Identity proofs I'm strugling with these two combinatorial identities:
$$  \binom{n+2}{3}=\sum_{i=1}^{n} i(n+1-i) $$
and
$$\binom{n+1}{2}^2=\sum_{i=1}^{n} i^3$$
Please give me some footholds and hints

For the first problem split the RHS:
$$\sum_{i=1}^n i(n+1-i) = \sum_{i=1}^n in +i - i^2 = n\sum_{i=1}^n i + \sum_{i=1}^n i - \sum_{i=1}^n i^2$$
Now we know that the some for consecutive integer from $1$ to $n$ is $\frac{n(n+1)}{2}$ and for the squares of consecutive integers from $1$ to $n$ is $\frac{n(n+1)(2n+1)}{6}$. So we substitute and we have:
$$n\sum_{i=1}^n i + \sum_{i=1}^n i - \sum_{i=1}^n i^2 = n \times \frac{n(n+1)}{2} + \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6}$$
$$n \times \frac{n(n+1)}{2} + \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)}{6} \times (3n + 3 - 2n - 1) = \frac{n(n+1)(n+2)}{3!} = \frac{(n+2)!}{3!(n-1)!} = \binom{n+2}{3}$$
And for the second problem it's a direct consequence from:
$$\sum_{i=1}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2 = \binom{n+1}{2}^2$$
Note that this substitutions for the summations are well known, but they are proven using induction.
