My statistics textbook states that $\sum\limits_{n=0}^{\infty} \begin{pmatrix}2n \\ n \end{pmatrix}(pqs^2)^n=\frac{1}{\sqrt{1-4pqs^2}}$
where $|s|<1$, $p \in ]0,1[$ and $q=1-p$.
That thing is that prof of this is omitted since its "not relevant". How would one prove it?
Soo far i have never calculated a sum involving a binomial coefficient.
It is a consequence of the generalized binomial theorem, for $|x|<1$, $$(1+x)^{-1/2}=\sum_{n=0}^{\infty}\binom{-1/2}{n}x^n,$$ where $$\binom{-1/2}{n}=\frac{(-1/2)(-1/2-1)\cdots(-1/2-(n-1))}{n!}=\frac{(-1)^n}{4^n}\binom{2n}{n}.$$ Hence for $|4pqs^2|<1$, $$\frac{1}{\sqrt{1-4pqs^2}}=(1+(-4pqs^2))^{-1/2}=\sum_{n=0}^{\infty}\binom{-1/2}{n}(-4pqs^2)^n\\=\sum_{n=0}^{\infty}\frac{(-1)^n}{4^n}\binom{2n}{n}(-4pqs^2)^n=\sum_{n=0}^{\infty}\binom{2n}{n}(pqs^2)^n.$$ Note that if $p\in(0,1)$ and $q=1-p$ then $pq\leq 1/4$ and for $|s|<1$ the condition $|4pqs^2|<1$ is satisfied.
Using Binomial series, for $|x|<1$ the $r(r\ge0)$th term of $$(1-x)^{-1/2}$$ is
$$\dfrac{-1/2(-1/2-1)\cdots\{-1/2-(r-1)\}}{r!}(-x)^r=\dfrac{1\cdot3\cdot5\cdots(2r-1)x^r}{r!2^r}=\binom{2r}r(x/4)^r$$
