Formula for $\sum_{i=1}^n (i *(i + 1))^2$ Is any formula for
$$\sum_{i=1}^n (i *(i + 1))^2$$ 
 A: Your sum may be written as $\sum_{i=1}^n i^4 + 2 \sum_{i=1}^n i^3 + \sum_{i=1}^n i^2$. There is a formula to compute those three sums, called Faulhaber's formula, see https://en.wikipedia.org/wiki/Faulhaber%27s_formula 
As you can see from the article, you need a bit of mathematical machinery to come up with this result (the binomial theorem, or Bernoulli numbers, or similar). However, without knowing anything about binomial coefficients or Bernoulli numbers or the like, you can cook up the Faulhaber formula for a given power $k$ by computing the interpolating polynomial of $(x,\sum_{i=1}^x i^k)$ for $x=0,1,\dots,k+1$. This does not prove that your result is correct for $x>k+1$, but that can be done by induction once you have the result in hand, if you like. (I wouldn't recommend it for $k=4$, though, the algebra will probably get nasty.) Similarly, since in your case you are just adding up three polynomials of degree at most $5$, you could compute your result by getting the interpolating polynomial of $(x,\sum_{i=1}^x (i(i+1))^2)$ for $x=0,1,\dots,5$.
A: You can use Faulhaber's relations: denoting $S_k(n)=1^k+2^k+\dots+n^k$ $\,(k\in\mathbf N$), you can prove with the binomial formula that
$$\sum_{k=0}^r\binom{r+1}{k} S_k(n)=(n+1)^{r+1} -1,$$
leading to a recursive computation of the $S_k(n)$.  Now
$$\sum_{i=1}^n\bigl(i(i+1)\bigr)^2=S_4(n)+2S_3(n)+S_2(n). $$
Further you may happen to already know that
$$S_2(n)=\frac{n(n+1)(2n+1)}{6}, \quad S_3(n)=\bigl(S_1(n)\bigr)^2=\frac{n^2(n+1)^2}{4}.$$
A: It is enough to exploit the hockey-stick identity. We have
$$\left[i(i+1)\right]^2 = 24\binom{i+2}{4}+4\binom{i+1}{2}\tag{1}$$
hence
$$ \sum_{i=1}^{N}\left[i(i+1)\right]^2 = 24\binom{N+3}{5}+4\binom{N+2}{3}=\frac{N(N+1)(N+2)(3N^2+6N+1)}{15}.\tag{2}$$
As an alternative, the closed form for the LHS of $(2)$ can be computed from Lagrange's interpolation, since $(2)$ certainly is a fifth-degree polynomial in the $N$ variable.
