Mistake in proof for sequence of cubes

Question

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...

The answer you got is right isn't it? I dont think there is any mistake here. — Win Vineeth, Mar 09 '16 at 13:58
Agreed: the answer you have is correct. It does factorise as $$\frac{n^2(n+1)^2}{4}$$ — Patrick Stevens, Mar 09 '16 at 14:02

score 6 · Accepted Answer · answered Mar 09 '16 at 14:01

6

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

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

answered Mar 09 '16 at 14:01

Banach Tarski

2,207

1

I am soooo retarded. Thanks a lot! – Mr Tsjolder from codidact Mar 09 '16 at 14:04

Mistake in proof for sequence of cubes

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