Power Series with complicated Algebra I need to find the Power Series for $\frac{ac^2}{\sqrt{1-\frac{b^2}{c^2}}}$. I said let $\frac{b^2}{c^2}=x$. Therefore we have $\frac{ac^2}{\sqrt{1-x}}$. Well we know the Power Series for $\frac{1}{1-x}$. Therefore we substitute and get
$$ac^2(1+x+x^2+x^3+x^4+\cdots)^{0.5}$$
If I substitute back in for $x$ though I get stuck on the next step. Any tips or other ways would be much appreciated.

Suppose
$f(x)
=\frac{1}{\sqrt{1-x}}
$.
Then
$f^2(x)
=\frac1{1-x}
=\sum_{n=0}^{\infty} x^{n}
$.
If
$f(x)
=\sum_{n=0}^{\infty} a_n x^n
$,
$\begin{array}\\
f^2(x)
&=\sum_{n=0}^{\infty} a_n x^n \sum_{m=0}^{\infty} a_m x^m\\
&=\sum_{n=0}^{\infty}  \sum_{m=0}^{\infty} a_n a_m x^{n+m}\\
&=\sum_{k=0}^{\infty}  \sum_{j=0}^{k} a_j a_{k-j} x^{k}\\
&=\sum_{k=0}^{\infty} x^k \sum_{j=0}^{k} a_j a_{k-j}\\
\end{array}
$
Therefore
$1
=\sum_{j=0}^{k} a_j a_{k-j}
$.
If $k=0$,
$1 = a_0^2$
so
$a_0 = 1$ or $-1$.
I will choose
$a_0 = 1$;
the other choice will
reverse all the signs.
If $k \ge 1$,
$1
=\sum_{j=0}^{k} a_j a_{k-j}
=2a_0a_k +\sum_{j=1}^{k-1} a_j a_{k-j}
=2a_k +\sum_{j=1}^{k-1} a_j a_{k-j}
$
or
$a_k
=\frac12(1-\sum_{j=1}^{k-1} a_j a_{k-j})
$.
For $k=1$,
$a_1 = \frac12
$.
For $k=2$,
$a_2
=\frac12(1-a_1^2)
=\frac12(1-\frac14)
=\frac{3}{8}
$.
As a check so far,
$(1+x/2+3x^2/8)^2
=1+x(1/2+1/2)+x^2(1/4+2(3/8)) + (...)x^3
=1+x+x^2+(...)x^3
$.
For $k=3$,
$a_3
=\frac12(1-a_1a_2-a_2a_1)
=\frac12(1-2a_1a_2)
=\frac12(1-2\frac12 \frac{3}{8})
=\frac{5}{16}
$.
And so on.
