Prove Summation $k=1$ to $n$ $k^3$ with telescoping rule I know how to do this problem when trying to get sum of squares 
$$\Sigma k^2  =  n(n+1)(2n+1)/ 6 $$
But I’m having trouble proving for cubes:
$$\sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}4$$
I have to prove this by the method of telescopy.
Edit $ $ Below is my attempt based on discussion on an answer below 
I started by  writing $\ \displaystyle \sum_{k=1}^n (k^4-(k-1)^4) = n^4.$
but I don't know where to go once I get here $\ \displaystyle \sum_{k=1}^n  (4k^3-6k^2+4k-1) = n^4 $ 
I used a table in a book for the others but I don't know how to convert the first  summand  $\ 4 \sum_{k=1}^n k^3$ to complete the proof.
Update $\ $ If anyone is interested in the exact solution I posted an answer showing what I did.
 A: Note that
$$
(k+1)^2k^2 - k^2(k-1)^2 = k^2[(k+1)^2 - (k-1)^2] = k^2[(2k)\cdot (2)] = 4k^3.
$$
Hence,
$$
4 \sum_{k = 0}^n k^3 = (n+1)^2n^2
$$
as wanted.
A: If anyone is interested in the exact solution here is what I did:

A: The proof reduces to simple polynomial arithmetic if you apply the telescopy theorem below, viz.
$$  \overbrace{\dfrac{n^2(n+1)^2}4}^{\Large F(n)}\, =\, \sum_{k\: =\: 1}^n\:\overbrace{k^3}^{\Large f(k)}  \iff \ \color{#c00}{F(1)=f(1)},\,\ \ F(n) - F(n\!-\!1)\ =\ f(n)\ \ {\rm for}\ \ n> 1$$
Checking: the base equation: $\color{#c00}{F(1)} = 1^2 2^2/4 = 1 = 1^3 = \color{#c00}{f(1)}\,$ is true, as is the inductive equation 
$$F(n)-F(n-1)\, =\, \dfrac{n^2(n+1)^2}4 - \dfrac{n^2(n-1)^2}4 = \dfrac{n^2(4n)}4 = n^3 = f(n)$$
That completes the proof by the Theorem below.

Theorem $\ $ (Additive Telescopic Induction)
$$ F(n)\ =\, \sum_{\large k\: =\: 1}^{\large n}\:\ f(k)\ \  \iff\ \ \ \color{#c00}{F(1)=f(1)},\,\ \ F(n) - F(n\!-\!1)\ =\ f(n)\ \ {\rm for}\ \ n> 1$$
Proof $\ (\Leftarrow)\ $ The $\,n=1\,$ base case $\,\color{#c00}{F(1) = f(1)} = \sum_{k=1}^1 f(k)\,$ is true, as is the inductive step:
$$\begin{align}  F(n\!+\!1)\ &=\ \  \color{#0a0}{F(n)}\ \  +\ f(n\!+\!1)\ \ \ {\rm by\ hypothesis}\\
&=\, \color{#0a0}{\sum_{i=1}^n f(i)} + f(n\!+\!1)\ \ \ {\rm by\ }\color{#0a0}{\rm induction}\\
&=\, \sum_{i=1}^{n+1} f(i)
\end{align}$$
That proves the nontrivial direction $(\Leftarrow)$. The reverse direction is clear.  $\ $ QED
Remark $ $ Note that the proof requires no ingenuity whatsoever. Rather it requires only simple computations that are so mechanical that they can be performed by a computer (namely equality testing of polynomials). In particular there is no need to pull your hair out in search of the appropriate inductive step - that is neatly encapsulated once and for all in the Theorem.
See this post and its links for much further discussion of telescopy.
A: Hint: If you already have an expression for the sum of consecutive squares then try playing with $\sum_{k=1}^n (k^4-(k-1)^4) = n^4.$
