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 $k^3$

  • $\begingroup$ $(k+1)^4-k^4=4k^3+6k^2+4k+1$ $\endgroup$ 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

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.

  • $\begingroup$ 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. $\endgroup$ 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 $k$. $\endgroup$ 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

\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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.