- #1

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^{n}).

(a+b)

^{n}= a

^{n}+ na

^{n-1}b + n(n-1)/2! a

^{n-2}b

^{2}+ ... + b

^{n}

My issue with this is that if one were to solve for n = 2, the b

^{2}term comes out from the 3rd term (2! term) and there is no need to add the b

^{n=2}term at the end.

Is my problem that if I am using n = 2, I shouldn't even bother to include the 3rd term (and all higher order terms that incidentally go to zero anyways)? It seems to me that the b

^{n}term is simply unnecessary at the end.

Thanks for any input!