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find the Mclaurin series for f(x) = (1-x)^(-2) So I took the derivative of f(x) = (1-x)^(-2) f(x) = (1-x)^(-2) f'(x) = 2(1-x)^(-3) f''(x) = 6(1-x)^(-4) f'''(x) = 24(1-x)^(-5) f''''(x) = 120(1-x)^(-6) then I took f(0) for all the derivatives f(0) = 1 f''(0) = 2 f'''(0) = 6 f''''(0) = 24 f'''''(0) = 120 Can anyone show me how to get a series from this?

Mathematics
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I think there is easier way but let me study it for a bit
I assume I use the formula \[\frac{f^{n}(0)(x)^{n}}{n!}\]
I will teach you a better way

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Other answers:

now , it is common knowlege(back of the book) \[\frac{1}{1-x}=\sum _{n=0}^{\infty } x^n\]
I need to use Mclaurin series, I know that method at least I think I do I still need to seek help to ensure I'm doing it correclty
though
you want to find series for \[\frac{1}{(1-x)^2}\]
yes but I have to use the Mclaurin series method
notice \[\frac{D}{\text{Dx}}\left[\frac{1}{(1-x)}\right]=\frac{1}{(1-x)^2}\]
I think @91 is leading you to the easiest method.
remember \[\frac{1}{(1-x)}=\sum _{n=0}^{\infty } x^n\] so take derivative of other side to get \[\frac{1}{(1-x)^2}=\sum _{n=0}^{\infty } n x^{n-1}\]
this question is more about learning how to use the Mclaurin series method than anything else
We should add that the convergence is only valid for |x|<1
I know the method he is using
Since the McLaurin Series is unique. It does not matter which method to use to find it.
ugh, I should have asked a question with a trig function or e^(x) something

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