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I am trying to solve this question but I'm not really understanding how to continue, I would greatly appreciate some kind of tip.

The Question

The formula for $1^2 + \cdots + n^2$ may be derived as follows. We begin with the formula $$(k+1)^3 - k^3 = 3k^2 + 3k + 1$$

Writing this formula for $k=1,\ldots,n$ and adding we obtain an expression.

Thus we can find $\sum_{k=1}^{n}k^2$ if we already know $\sum_{k=1}^n k$.

Use this method to find $1^3 + \cdots + n^3$.

I am confused about this method. How did they get the formula to begin with, and what formula am I supposed to use for $1^3 + \cdots + n^3$?

EDIT: I noted that they did already hint at $k^2$ being expressed from $k^1$, but we can't square $k^1$ into $k^3$

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  • $\begingroup$ $(k+1)^4-k^4=4k^3+6k^2+4k+1$ $\endgroup$ – André Nicolas Jun 18 '14 at 5:58
  • $\begingroup$ How did you get that expression? I noted that you basically did $(k+1)^{(3+1)}$ but what reason would there be? $\endgroup$ – Jason Jun 18 '14 at 6:02
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The formula is the Binomial Theorem, or just multiplication: $(k+1)^3=k^3+3k^2+3k+1$. This can be rewritten as $$(k+1)^3-k^3=3k^2+3k+1.$$

The corresponding formula for fourth powers is $(k+1)^4=k^4+4k^3+6k^2+4k+1$. It yields the identity $$(k+1)^4-k^4=4k^3+6k^2+4k+1.$$ That can be used to give a telescoping argument for the sum of the first $n$ cubes, much like the telescoping argument for the sum of the first $n$ squares. Sum both sides from $k=1$ to $k=n$. On the right, there is almost total cancellation, and we get $$(n+1)^4-1=4\sum_1^n k^3+6\sum_1^n k^2+4\sum_1^n k +\sum_1^n 1.\tag{1}$$ Now a fair bit of messy algebra gets us $\sum_1^n k^3$, since we have formulas for every other sum in (1).

In general, the Binomial Theorem is the assertion that if $m$ is a positive integer, then
$$\small (x+y)^m=\binom{m}{0}x^m +\binom{m}{1}x^{m-1}y+\binom{m}{2}x^{m-2}y^2+\cdots +\binom{m}{m-1}xy^{m-1}+\binom{m}{m}y^m.$$

Remark: It will turn out that $$1^3+2^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2.$$ Once we have guessed this formula, it can be proved more easily by induction.

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  • $\begingroup$ Couldn't we just skip all of the above steps and start with $\sum_j^n j^3$ because it is logical to do so? It technically is the representation of the cubes of the sums from j = 1 to j = n $\endgroup$ – Jason Jun 19 '14 at 1:52
  • $\begingroup$ We are looking for a closed form expression for $\sum_1^n j^3$. The trick using $(k+1)^4-k^4$ is one way of getting such a closed form. $\endgroup$ – André Nicolas Jun 19 '14 at 1:56
  • $\begingroup$ Isn't expressing $k^3$ as $(k+1)^4 - k^4 = 4k^3 + 6k^2 + 4k + 1$ incorrect? Say k = 3, k^3 = 27, but the above expression gives 256 - 81 = 108+54+12+1. There is no way to add any of this together to get 27. $\endgroup$ – Jason Jun 19 '14 at 2:10
  • $\begingroup$ Note that $256-81=175$ while $108+54+12+1=175$, so the equation is correct when $k=3$. It is in fact correct for all $k$. $\endgroup$ – André Nicolas Jun 19 '14 at 2:15
  • $\begingroup$ Yes, but doesn't that have nothing to do with k^3? i.e. it has no relation to the sum of k^3 from i = 1 to n $\endgroup$ – Jason Jun 19 '14 at 2:19
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\begin{align} 1^3-0^3=3*0^2+3*0+1\\2^3-1^3=3*1^2+3*1+1\\...\\(n+1)^3-n^3=3*n^3+3*n+1 \end{align} Now sum all of the equalities. \begin{align} (n+1)^3=3\sum_{k=1}^{n}k^2+3\sum_{k=1}^{n}k+(n+1) \end{align} Now you can proceed further.

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