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I trying to work through a problem and have become stuck at the following equality:

$ \sum_{n=1}^{100}{ n^2+n - 1 - (n-1)^2} = \sum_{n=1}^{100}{(3n - 2)}$

I can't quite get my head around the factoring. Could anybody help out? Many thanks!

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$(n-1)^2=n^2-2n+1$ –  pedja Apr 25 '12 at 6:33
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2 Answers 2

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It isn't factoring, it is unfactoring (expanding). We have $(n-1)^2=n^2-2n+1$ and therefore $$(n^2+n-1)-(n-1)^2=(n^2+n-1)-(n^2-2n+1)=3n-2.$$

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Since: $$(n-1)^2=n^2-2n+1$$ Then: $$n^2+n-1-(n-1)^2=n^2+n-1-(n^2-2n+1)=3n-2$$

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