2
$\begingroup$

I know there are thousands of proofs for this to have a look at, but I started one myself in a slightly different way than what is easily found when googling. To me, the proof seems like it should work, but the result is not what it should be

$$\begin{align} \sum_{k=0}^n k^3 & = \sum_{k=0}^n (n-k)^3 \\ & = \sum_{k=0}^n (n^3 -3n^2k + 3nk^2 - k^3) \\ & = (n+1)n^3 - 3n^2\frac{n(n+1)}{2} + 3n\frac{n(2n+1)(n+1)}{6} - \sum_{k=0}^n k^3\\ \Rightarrow 2 \sum_{k=0}^n k^3 & = \frac{2n^3(n+1)}{2} - \frac{3n^3(n+1)}{2} + \frac{(2n^3 + n^2)(n+1)}{2} \\ \Rightarrow \sum_{k=0}^n k^3 & = \frac{(n^3 + n^2)(n+1)}{4} \end{align}$$

I realise it might be easier or something like that when considering a sum starting from $k=1$ (that is what everyone seems to be doing in any case), but I thought it should work like this as well. Did I make any mistake throughout my proof or forget something? I assume it is provable like this...

$\endgroup$
  • $\begingroup$ The answer you got is right isn't it? I dont think there is any mistake here. $\endgroup$ – Win Vineeth Mar 9 '16 at 13:58
  • 2
    $\begingroup$ Agreed: the answer you have is correct. It does factorise as $$\frac{n^2(n+1)^2}{4}$$ $\endgroup$ – Patrick Stevens Mar 9 '16 at 14:02
5
$\begingroup$

You are correct. Just take $n^2$ common from the first parenthesis.

$$\frac{(n^3 + n^2)(n+1)}{4} = \frac{n^2(n+1)^2}{4} = \left(\frac{n(n+1)}{2}\right)^2$$

$\endgroup$
  • 1
    $\begingroup$ I am soooo retarded. Thanks a lot! $\endgroup$ – Mr Tsjolder Mar 9 '16 at 14:04

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.