Find a formula for $\sum_{k=-3}^n$ $k^3 (k + 1) (k + 2) (k + 3)$ 
Find a formula for $\sum_{k=-3}^n$ $k^3 (k + 1) (k + 2) (k + 3)$ and prove it.

I'm pretty sure that once I find a formula I can prove it by induction. I'm just struggling to find the formula. My professor hasn't really taught us how to do that part. I can plug numbers in to get an idea, but I can't seem to formulate a formula. Suggestions?
For $n=1$, the result is 24. For $n = 2$, the result is 504. I'm guessing the formula will involve factorials.
 A: Here’s a method of solving it. It will take a little bit of work (algebra), but you should be able to do it.
Let $S(n)$ be the value of this sum, in terms of the upper limit $n$. Then you know that
$$\begin{align}S(n) - S(n-1) &= n^3(n+1)(n+2)(n+3) \\&= n^3(n^3 + 6n^2 + 11n + 6) \\&= n^6 + 6n^5 + 11n^4 + 6n^3\end{align}$$
There are infinitely many polynomials in $n$ satisfying this difference equation, and they differ by a constant. It turns out that $S(n)$ can be uniquely expressed as one of these seventh-degree polynomials:
$$S(n)=x_7n^7 + x_6 n^6 + ...+x_1 n + x_0$$
You just need to figure out the values of these eight coefficients. To do this, you can treat the following sequence of identities as a system of eight equations with eight unknowns $x_i$:
$$0x_7 + 0x_6 + ... + 0x_1 + 1x_0 = S(0) \\ 1x_7 + 1x_6 + ... + 1x_1 + 1x_0 = S(1) \\ 2^7 x_7 + 2^6 x_6 + ... + 2x_1 + 1x_0 = S(2) \\ ... \\ 7^7 x_7 + 7^6 x_6 + ... + 7x_1 + 1x_0 = S(7)$$
You can solve this system easily with a bit of linear algebra, but you probably don’t want to do it by hand. And once you find out the values of the $x_i$, you will know the coefficients of the polynomial $S(n)$.
A: Note
$$
\begin{align*}
k^3 &= k^3-1+1 \\
    &= k(k-1)(k+1) + 1 \\
    &= k(k-1)(k-2) + 3k(k-1) + 1 
\end{align*}
$$
So given is
$$
\sum (k-2)(k-1)k(k+1)(k+2)(k+3) + 3{\sum(k-1)k(k+1)(k+2)(k+3)} + \sum (k+1)(k+2)(k+3) $$
$$ = (1\cdot2\cdot3\cdot4\cdot5\cdot6 + 2\cdot3\cdot4\cdot5\cdot6\cdot7 + \ldots) + 3(1\cdot2\cdot3\cdot4\cdot5 + 2\cdot3\cdot4\cdot5\cdot6 + \ldots) + (1\cdot2\cdot3 + 2\cdot3\cdot4 + \ldots) $$
Can you end it now?
(Careful about the new bounds)
