Find a closed form for the sum $∑(x^3 - 2x)$ from $x=1$ to any number $n$ Find a closed form for the sum $∑(x^3 - 2x)$ from $x=1$ to any number $n$.
Can someone explain to me what a closed form is and how to approach this problem?
 A: Just like integration, summing the values a polynomial expression gives you a polynomial expression one degree higher. So, write $\sum_{x=1}^n x^3-2x = a_4n^4 + a_3n^3+a_2n^2+a_1n+a_0$ and set $n=1,2,3,4,5$ to find the coefficients by solving a $5\times 5$ linear system.
Or you can compute the first 5 values and all differences:
-1   3  24   80   195
 4  21  56  115
17  35  59
18  24
6

and get the formula
$$
-1{n-1 \choose 0} + 4{n-1 \choose 1} + 17{n-1 \choose 2} + 18{n-1 \choose 3} + 6{n-1 \choose 4} = \frac14 n (n+1) (n^2+n-4)
$$
A: A closed form is an expression that doesn't involve any looping/repeatition/recursion (like sums, products, sequences, etc.). It is not exactly clear, for example some consider binomials $\binom{n}{k}$ a closed form expression, while others do not. However, in case of sums of polynomials, it is quite clear: it is another polynomial. For example $\sum_{k=1}^{n}k = \frac{n(n+1)}{2} = \frac{1}{2}n^2+\frac{1}{2}n$.
In your case we will try to calculate 
$$F(n) = \sum_{k=1}^{n} k^3-2k.$$
Observe, that $P(k) = k^3-2k$ is a polynomial of degree $3$, and so, its sum will be of degree $4$ (this is closely related to the fact that the integral of polynomial of degree $d \in \mathbb{N}$ is of degree $d+1$).
Now, take an arbitrary polynomial of degree $4$:
$$F(n) = a_4n^4+a_3n^3+a_2n^2+a_1n+a_0$$
Note, that it has $5$ unknowns, namely $a_4,a_3,\ldots,a_0$. This allows us to take some $5$ values of $F$ calculated by hand and obtain a linear set of equations (here I started with $-2$ to make calculations easier, but starting with $0$ or any other number is ok too):
\begin{align*}
\begin{cases}
F(-2) &= -1 \\
F(-1) &= 0 \\
F(0) &= 0 \\
F(1) &= -1 \\
F(2) &= 3 \\
\end{cases}
\end{align*}
We could solve this right away, but it would be tedious. To make it shorter, note the two zeros at $0$ and $-1$, this means $F(n) = n(n+1)G(n)$ where $G$ is of degree $2$. Some simple manipulations (i.e. $G(n) = \frac{F(n)}{n(n+1)}$) get us
\begin{align*}
\begin{cases}
G(-2) &= \dfrac{-1}{(-2)\cdot(-1)} &= -\dfrac{1}{2} \\
G(1) &= \dfrac{-1}{1\cdot2} &= -\dfrac{1}{2} \\
G(2) &= \dfrac{3}{2\cdot3} &= \dfrac{1}{2} \\
\end{cases}
\end{align*}
Solving the above is easy, we get $G(n) = \frac{n^2}{4}+\frac{n}{4}-1$, and so $$F(n) = n(n+1)\left(\frac{n^2}{4}+\frac{n}{4}-1\right).$$
I hope this helps $\ddot\smile$
