Binomial expansion with only odd coefficients?

In William Feller's 1st book p.272

It said the generating function $\Phi$ satisfies

\begin{equation*} qs\Phi^2(s) - \Phi(s) + ps = 0 \end{equation*}

so it has two roots. The first root is unbounded near $s = 0$.

So the generating function is given by the unique bounded solution

\begin{equation*} \Phi(s) = \frac{1 - \sqrt{1 - 4pqs^2}}{2qs}. \end{equation*}

Also, \begin{equation*} \Phi(s) = \sum_{n=0}^\infty \phi_ns^n. \end{equation*}

What I don't understand is how the coefficients \begin{equation*} \phi_{2k-1} = \frac{(-1)^{k-1}}{2q} \binom{1/2}{k} (4pq)^k, \quad \phi_{2k} = 0 \end{equation*} come up with binomial expansion of the unique root.

To be clear, I want to know how the unique root of $\Phi$ is converted to the form $(1 + t)^a$ and how and why the new form have the coefficients above?

-
What does this tell you about the symmetry of $\Phi(s)$? Make a plot... – draks ... Jul 31 '12 at 7:08
Could you elaborate? – RHS Jul 31 '12 at 7:12
Compare the symmetry of $x^{2n}$ resp. $x^{2n+1}$, when you substitute $-x$ for $x$. What do you get? – draks ... Jul 31 '12 at 7:15
The explicit formula for the coefficients $\phi_k$ follows from the expansion of $(1+x)^{1/2}$ for $x=-4pqs^2$. The coefficient of $x^k$ in $(1+x)^{1/2}$ is ${1/2\choose k}$, which yields the coefficient of $s^{2k-1}$. – Did Jul 31 '12 at 7:20
Google Binomial series. – Did Jul 31 '12 at 7:32

First, by Newton's binomial formula we have \begin{equation*} (1 - 4pqs^2)^{1/2} = \sum_{k=0}^\infty \binom{1/2}{k} (-4pqs^2)^k. \end{equation*} And its first term $\binom{1/2}{0} (4pqs^2)^0$ is $1$. So \begin{equation*} \Phi(s) = \frac{1 - \sqrt{1 - 4pqs^2}}{2qs} = - \frac{1}{2qs} \sum_{k=1}^\infty \binom{1/2}{k} (-4pqs^2)^k = \frac{(-1)^{k+1}}{2q} \sum_{k=1}^\infty \binom{1/2}{k} (4pq)^k s^{2k -1}. \end{equation*} Now we can see $\Phi$ only has odd coefficient, and we could get $\phi_{2k-1}$ and $\phi_{2k}$ immediately.