Calculating Coefficients Using Geometric Series

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SUMMARY

The discussion centers on calculating the coefficients of the expression [x^n] (1-2x+x^2)^{(-k)} using geometric series. The transformation of the expression to [x^n] \frac{1}{(x-1)^{(2k)}} is established as a key step. Participants suggest rewriting the expression in terms of a geometric series and differentiating term by term to find the coefficients. The notation [x^n]f(x) is clarified as denoting the coefficient of x^n in the function f(x).

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Homework Statement


Calculate [tex][x^n] (1-2x+x^2)^{(-k)}[/tex]

Homework Equations


Just the geometric series.

The Attempt at a Solution


This is what I got so far [tex][x^n] (1-2x+x^2)^{(-k)} = [x^n]((x-1)^2)^{(-k)} = [x^n] \frac{1}{(x-1)^{(2k)}}[/tex]
Basically, how do I put this into a sum form? or can I just multiply like this [tex]\frac{1}{(x-1)^{(2k)}} * \frac{1-x}{1-x} \Rightarrow a_n = \frac{(1-x)}{(x-1)^{(2k)}}[/tex]?
 
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I have no clue what you are asking! "Calculate ...". What do you mean by "calculate" an expression? You say "just the geometric series". What about a geometric series? Are you asked to write that expression in terms of a geometric series?
 
Differentiate term by term 2k times:

[tex]\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots[/tex]

NB: Halls, [x^n]f(x) is sometimes used to denote the coefficient of x^n in f(x) .
 

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