Find a power series representation for this function

Question

Find a power series representation centered at $x=0$ for $f(x) = \dfrac x {15x^2+1}$.

Answer: $f(x) = \sum \limits _{n=0} ^\infty (-1)^n (15x^2)^{n+1}$.

For some reason this just isn't clicking with me haha.

Here is my thought process:

$f(x) = \dfrac{x}{15x^2 + 1} $

$f(x) = x \times \dfrac1{1 + 15x^2} $

get the top x out of there and switch their positions around in the denominator

$f(x) = x \times\dfrac 1{1 - (-15x^2)} $

now it is in the form of $\dfrac1{1 -x}$ (my $x$ being a $-15x^2$)

$$x \sum_{n=0}^\infty (-1)^n \times (15x^2)^n $$

And here is where I have no idea what to do

Answer 1

It should be $$x\sum_{n=0}^\infty (-1)^n 15^n x^{2n}=\sum_{n=0}^\infty (-1)^n 15^n x^{2n+1}$$

and not $$\sum_{n=0}^\infty (-1)^n (15 x^{2})^{n+1}$$

Answer 2

Rewrite the given function in the form of $$\frac{a}{1-r}$$ so you can express the representation as a geometric series. A geometric series takes the form of $$ \sum_{n=0}^{\infty } a(r)^n$$

Answer 3

$$\frac{x}{15x^2+1}=\frac{x}{1+(\sqrt{15}x)^2}$$ now we can depend on the geometric series which is $$\frac{1}{1+x}=\sum_{n=0}^{\infty }(-1)^nx^n$$ let $x\rightarrow x^2$ $$\frac{1}{1+x^2}=\sum_{n=0}^{\infty }(-1)^nx^{2n}$$ to get the power series let $x\rightarrow \sqrt{15}x$,so $$\frac{x}{1+(\sqrt{15}x)^2}=x\sum_{n=0}^{\infty }(-1)^n(\sqrt{15}x)^{2n}=x\sum_{n=0}^{\infty }(-1)^n(15x^2)^n=\sum_{n=0}^{\infty }(-1)^n(15x^2)^n.x$$