TITLE: Find closed formula without using induction for $\sum_{k=0}^n k^3$
QUESTION [6 upvotes]: I'm studying for my exam in discrete mathematics and found the following problem on last years exam:
Find a closed formula without using induction for $\sum_{k=0}^n k^3$.
I tried it by finding the Generating Function first:
$F(x) = F_0 + \sum_{k=1}^nF_nx^n = \sum_{k=1}^n (F_{n-1}+n^3)x^n = \sum_{k=1}^n F_{n-1}x^n + n^3x^n = \sum_{k=0}^n F_n x^{n+1} + \sum_{k=1}^n n^3x^n = xF(x) + \sum_{k=1}^nn^3x^n$
The problem seems to be, that I lack an actual recursive definition of $\sum_{k=0}^n k^3$ which is, as far as I know, needed to find a generating function. Above, I pretty much used, that $X_n = X_{n-1}+n^3$, but obviously, that`s not enough. Because a recursive definition was always given in our lectures, I don't now other possibilities to solve this, except for finding the Generating Function with help of recursive definitions.

REPLY [2 votes]: To use a generating function to solve this, it helps to know that the ordinary generating function for the sequence $\{n^m\}_{n=0}^\infty$ is $\left(x\frac d{dx}\right)^m\frac1{1-x}$: differentiation multiplies the coefficient of $x^n$ by $n$ and shifts the sequence down, and multiplying by $x$ shifts back up and sets the constant term to $0$, so the net effect is to multiply the $x^n$ term by $n$.  
Following the recipe for finding an ordinary generation function, we start with the recurrence $$a_n = a_{n-1}+n^3, n\gt0, a_0=0.$$ Multiply both sides by $x^n$ and sum over valid values of $n$: $$\sum_{n=1}^\infty a_n x^n = \sum_{n=1}^\infty a_{n-1}x^n + \sum_{n=1}^\infty n^3x^n \\ \sum_{n=0}^\infty a_n x^n = x\sum_{n=0}^\infty a_nx^n + \sum_{n=0}^\infty n^3x^n \\
g(x) = x g(x) + \left(x \frac d{dx}\right)^3 \frac1{1-x} \\
g(x) = {x+4x^2+x^3 \over (1-x)^5}.$$ 
You could’ve instead arrived at this directly with an operator-based approach. As explained above, the “multiply by $n$” operator is $x\frac d{dx}$, while if $f(x)$ is the o.g.f. for a sequence, then $f(x)/(1-x)$ is the o.g.f. for the sequence of its partial sums. Starting from the o.g.f. for the sequence of all $1$s, $1/(1-x)$, the o.g.f. for the sequence of sums of the first $n$ cubes is therefore $$g(x) = \frac1{1-x}\left(x\frac d{dx}\right)^3\frac1{1-x}.$$ 
I’ll leave extracting the coefficient of $x^n$ from this to you.