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Proof for formula for sum of sequence 1+2+3+…+n?

I have this sigma:$$\sum_{i=1}^{N}(i-1)$$

is it $$\frac{n^2-n}{2}\quad?$$

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Why the tag "measure theory"? Do you know the formula $\sum_{j=1}^nj=\frac{n(n+1)}2$? You should write $N$ instead of $n$ in the final formula. – Davide Giraudo Oct 2 '11 at 19:29
sorry, I've wrote sigma then inter in the tag and I didn't know it was measure theory :S , yes I know this formula and wt I did is subtract N from it then I got what I wrote in the question,is it right? – Sosy Oct 2 '11 at 19:35
@Sosy: Yes, that is a correct argument. I've closed this question as a duplicate, because this question has come up before here. – Zev Chonoles Oct 2 '11 at 19:40
ok thanks a lot – Sosy Oct 2 '11 at 19:41

marked as duplicate by Hans Lundmark, Zev Chonoles Oct 2 '11 at 19:39

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There is an elementary proof that $\sum_{i = 1}^n i = \frac{n(n+1)}{2}$, which legend has is due to Gauss. For a proof, see my blog post at Math ∩ Programming.

What you have is the same as $\sum_{i = 1}^{N-1} i$, since adding zero is trivial. So your thing is correct, and substituting $n$ for $N-1$ we get $N(N-1)/2$, which is what you have.

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In fact, the legend does not say Gauss invented the formula (he didn't), but just that Gauss re-discovered it as a schoolboy. – GEdgar Oct 2 '11 at 19:42
Right. I meant the particular proof is due to Gauss (which might also be untrue). – Bean Oct 2 '11 at 20:48

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