Taylor series representation for $$ \frac{x}{(1+4x)^2}$$

[The Subject]

Homework Statement

Find a power series that represents $$ \frac{x}{(1+4x)^2}$$

Homework Equations

$$ \sum c_n (x-a)^n $$

The Attempt at a Solution

$$ \frac{x}{(1+4x)^2} = x* \frac{1}{(1+4x)^2} $$
since [tex] \frac{1}{1+4x}=\frac{d}{dx}\frac{1}{(1+4x)^2} [/tex]
$$ x*\frac{d}{dx}\frac{1}{(1+4x)^2} =x\frac{d}{dx}\sum_{n=0}^\infty(-4)^nx^n=x\sum_{n=0}^\infty(-4)^nnx^{n-1}=\sum_{n=0}^\infty(-4)^nnx^{n}$$

The solution suggests $$\sum_{n=0}^\infty(-4)^n(n+1)x^{n+1}$$

Am i doing something incorrect?

 

[fresh_42]

Reconsider your differentiation. Isn't ##\frac{d}{dx} x^{-2} = -2x^{-3}##?