Can someone explain to me why this is equal?
$$\sum_{i = 1}^n i = \sum_{i = 1}^n (n - i + 1) = \sum_{i = 0}^{n - 1} (n - i)$$
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What do you get when you try it with $n=5$? – GEdgar Jan 28 '15 at 18:32
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In the left most you sum $;1+2+3+\ldots+n;$ , in the central one you sum $;n+(n-1)+\ldots+3+2+1;$ , and the rightmost one is just a nicer form to write the central one. – Timbuc Jan 28 '15 at 18:34
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If you write out their sums:
The first one is $1+2+\dotsb+(n-1) + n$.
The second one is $n + (n-1) + \dotsb +2+1$.
The third one is again $n + (n-1) + \dotsb +2+1$.
Mike Pierce
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Oh so the second and third were just backwards, but it still adds up the same, thanks a bunch :) – kalvin Jan 28 '15 at 18:36
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@kalvin, yeah if you ever have a question about summations or sequences, it's never a bad idea to just start writing out the terms. – Mike Pierce Jan 28 '15 at 18:47