Lets see here...
$$f(x)=\frac{x^2}{(1-2x)^2}$$
$$=x^2\frac{d}{dx}\frac{1}{(1-2x)}$$
$$=x^2\frac{d}{dx}\left(\sum_{n=0}^{\infty}(2x)^n\right)$$
$$=x^2\sum_{n=0}^{\infty}2nx^{n-1}$$
$$=\sum_{n=0}^{\infty}2nx^{n+1}$$

Question

Lets see here...
$$f(x)=\frac{x^2}{(1-2x)^2}$$
$$=x^2\frac{d}{dx}\frac{1}{(1-2x)}$$
$$=x^2\frac{d}{dx}\left(\sum_{n=0}^{\infty}(2x)^n\right)$$
$$=x^2\sum_{n=0}^{\infty}2nx^{n-1}$$
$$=\sum_{n=0}^{\infty}2nx^{n+1}$$

anonymous · Answer

how did I do

anonymous · Answer

a little mistake in derivating...

anonymous · Answer

pretty sure u missed a coeff

anonymous · Answer

yeah lol

anonymous · Answer

multiply by -.5 and u should be good!

anonymous · Answer

and we must have the condition  $|2x|<1$

anonymous · Answer

good catch.

anonymous · Answer

actually not just a coefficient...

anonymous · Answer

lol yeah her second differentiation is wrong too...

anonymous · Answer

and when you fix that, I don't think it will simplify nicely.

anonymous · Answer

@MathSofiya 

note that  $$\frac{d}{dx} (2x)^n=2n(2x)^{n-1} $$

anonymous · Answer

or at least not without havving a clunky 2^n

anonymous · Answer

$$\sum_{n=0}^{\infty}2n(2x)^{n-1}$$ what should I do about the x^2?

anonymous · Answer

well as i said
if u distribute the n-1 exp over the 2 and the x

anonymous · Answer

then u can have a 2^n term and ur x^n+1 term

anonymous · Answer

$$f(x)=\frac{x^2}{(1-2x)^2}$$
$$=\frac{x^2}{2}\frac{d}{dx}\frac{1}{(1-2x)}$$
$$=\frac{x^2}{2}\frac{d}{dx}\left(\sum_{n=0}^{\infty}(2x)^n\right) \ \ \ \ if \ \ \ |2x|<1$$
$$=\frac{x^2}{2}\sum_{n=0}^{\infty}2n(2x)^{n-1}$$
$$=\sum_{n=0}^{\infty}n2^{n-1}x^{n+1}$$

anonymous · Answer

$$\sum_{n=0}^{\infty}2n2^{n-1}x^{n+1}$$

anonymous · Answer

well yes but as i said
 u are missing a -.5 and the 2 times 2^n-1 is just 2^n

anonymous · Answer

thank you

anonymous · Answer

yeah np

anonymous · Answer

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