Help! find the taylor series for f(x) center at the given value

Question

el_arrow · Answer

f(x)=9x^-2 and a=1

el_arrow · Answer

i already got the series i just need the sum |dw:1437185617267:dw|

el_arrow · Answer

@triciaal @Michele_Laino

el_arrow · Answer

i got 9 - 18/1! (x-1) + 52/2! (x-1)^2 - 216/3! (x-1)^3 + 1080/4! (x-1)^4 - 6480/5! (x-1)^5

el_arrow · Answer

@dumbcow

el_arrow · Answer

@Concentrationalizing help with one more problem

anonymous · Answer

What do you mean? You have the series but need the sum? Saying you need a sum implies that there's a numerical answer when there isnt.

anonymous · Answer

You mean you need the nth term?

el_arrow · Answer

yeah thats what i meant

anonymous · Answer

I would suggest doing what we did before, actually. Consider integrating the function and then applying the new center. 
$$\int\limits_{}^{}\frac{ 9 }{ x^{2} }dx = -\frac{ 9 }{ x }+C$$
Now I want to shift this antiderivative. It would have been too difficult to do so while it was squared, though. So what we can do is write this in the form of a geo-series by doing this:
$$-\frac{ 9 }{ x } + C = \frac{ 9 }{ 1-x-1 } = \frac{ 9 }{ 1-(x+1) } = 9\sum_{n=0}^{\infty} (x+1)^{n} + C$$ 
Now proceed exactly like we did in the last problem. Since we'll be differentiating, we don't need to worry about the +C.

el_arrow · Answer

okay so that would be (9x+1)^n right

anonymous · Answer

No, the 9 is just stuck outside. Not much you can do with it. Gotta head out, though. Good luck :)

el_arrow · Answer

@iambatman could you please help?