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Hello I was trying to find the coefficient for the member $x^5$ for the expansion: $(1-2x)^{-2}$

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up vote 3 down vote accepted

Hint: You know the expansion of $\dfrac{1}{1-t}$. Differentiate.

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Around $x=0$, we have that $$\dfrac1{1-2x} = 1 + (2x) + (2x)^2 + (2x)^3 + (2x)^4 + (2x)^5 + (2x)^6 + \cdots = \sum_{k=0}^{\infty} 2^kx^k$$ Hence, $$\dfrac{d}{dx} \left(\dfrac1{1-2x} \right)= \dfrac2{\left(1-2x \right)^2} = \sum_{k=1}^{\infty} k2^k x^{k-1}$$ Hence, $$\dfrac1{\left(1-2x \right)^2} = \sum_{k=1}^{\infty} k2^{k-1} x^{k-1}$$ Now you can read off the power of $x^n$ from above.

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Thank you very much @Marvis – Grangj Oct 20 '12 at 18:28

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