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Homework Statement
Using that [itex]\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n [/itex] for |x|<1 and that
[itex] f'(x) =\sum_{n=0}^{\infty} (n+1)a_{n+1}(x-x_0)^n [/itex], write [itex] \sum_{n=0}^{\infty} n^2x^n [/itex] in closed form.
Yagoda said:I am considering f(x) to be the closed form of the series that I'm looking for so I use the theorem to find its derivative f'(x) and then hopefully find f itself. So what I've got so far is that given f'(x), [itex]f(x) =\sum_{n=0}^{\infty} (n+1)^2(x)^n [/itex], which tell us that [itex]f(x) =\sum_{n=0}^{\infty} (n+1)^2(x)^n =\sum_{n=0}^{\infty} n^2x^n [/itex] since that was the original series, if that makes sense.
Maybe this is an elementary question, but is there a way to apply the geometric series identity to your expression to get a closed form or does some other type manipulation that I'm not seeing have to be done?