Prove Summation $k=1$ to $n$ $k^3$ with telescoping rule

Question

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.

Answer 1

Hint: If you already have an expression for the sum of consecutive squares then try playing with $\sum_{k=1}^n (k^4-(k-1)^4) = n^4.$

Answer 2

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.

Answer 3

If anyone is interested in the exact solution here is what I did:

Answer 4

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.

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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.