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