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I was studying the iteration method and one sample had this summation:

$\sum_{j=0}^{n-2} (n - j)$

Where this eventually gets simplified as:

$n(n-1) - {(n-2)(n-1)\over 2} $

I did not quite understand how this is achieved. The first step is obviously split them so:

$ \sum_{j=0}^{n-2} n - \sum_{j=0}^{n-2} j$

but what about the rest?

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    $\begingroup$ It's the old summation formula $\sum_{j=0}^k j = \frac{k(k+1)}{2}$ that Gauss found as a kid (it was of course known before that to mathematicians). $\endgroup$ – Daniel Fischer Aug 13 '13 at 10:42
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In the first sum we add $n-1$ times the number $n$ so we find $n(n-1)$ and for the second sum we can prove the result by induction or we use the Gauss method: denote by $S$ the desired sum so $$1+2+\cdots+(n-2)=S\\ (n-2)+(n-1)+\cdots 1=S$$ then if we add the first term with the first in the two equalities and the second with the second and so on we find $$(n-1)+(n-1)+\cdots+(n-1)=(n-2)(n-1)=2S$$ so we deduce $S$.

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  • $\begingroup$ "In the first sum we add n−1 times the number n" but why? How do we come to that solution? Also for the second one why do we wan the 2S?I did not quite understand the second part too... Could you explain a bit more? $\endgroup$ – Sarp Kaya Aug 13 '13 at 10:54
  • $\begingroup$ For the second sum we add the two equalities term by term and on the RHS we add $S+S=2S$ Right? $\endgroup$ – user63181 Aug 13 '13 at 10:59
  • $\begingroup$ Yes, I think I got it... $\endgroup$ – Sarp Kaya Aug 13 '13 at 11:01
  • $\begingroup$ Also I have another question, what about j equalities? Let's say j=2 instead of j=0 $\endgroup$ – Sarp Kaya Aug 13 '13 at 11:09
  • $\begingroup$ In the sum $\sum_{j=p}^q$ the number of terms is $q-p+1$. $\endgroup$ – user63181 Aug 13 '13 at 11:20
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You can change the order of summation and let variable $k=n-j.$ So $$\sum_{j=0}^{n-2} (n - j)$$ becomes $$\sum_{k=2}^{n} k$$ which is $\frac{n(n+1)}2-1$

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