How to write $\left(\frac{A+Bs}{C+Ds}\right)^N$ as $\sum_{n=-\infty}^{\infty} s^n P_n$? We are given this thing
$$J=\left(\frac{A+Bs}{C+Ds}\right)^N$$
Where $A,B,C,$ and $D$ are non-zero constants, $N$ is a positive constant.
We are told to find $P_0$ (the coefficient of the term $s^0$ in the sum) if
$$\left(\frac{A+Bs}{C+Ds}\right)^N=\sum_{n=-\infty}^{\infty}s^n P_n$$
And we know that $\forall n \notin \{0,1,\dots,N\}; P_n=0$.
We can say
$$\left(\frac{A+Bs}{C+Ds}\right)^N=\left(\frac{\tilde A+\tilde Bs}{1+\tilde D s}\right)^N$$
Where $\tilde X=X/C$.
I know that if we can write $J$ as $(\alpha s^i + \beta s^j)^N$, then we can use the binomial theorem to write:
$$J=\sum_{n=0}^{N} \frac{N!}{n!(N-n)!} (\alpha s^i)^n (\beta s^j)^{N-n}$$
$$=\sum_{n=0}^{N} s^{ni+(N-n)j}\frac{N!\alpha^n \beta^{N-n}}{n!(N-n)!}$$
$$=\sum_{n=0}^{N} s^{Nj+(i-j)n}\frac{N!\alpha^n \beta^{N-n}}{n!(N-n)!}$$
So $P_{Nj+(i-j)n}=\frac{N!\alpha^n \beta^{N-n}}{n!(N-n)!}$.
$$Nj+(i-j)n=0 \implies n={\frac{Nj}{j-i}}$$
$$\implies P_0=\frac{N!}{{\frac{Nj}{j-i}}!(N-{\frac{Nj}{j-i}})!} \alpha^{\frac{Nj}{j-i}} \beta^{N-{\frac{Nj}{j-i}}}$$
But I cannot write $J$ in that form.
How can I find $P_0$, the coefficient of the term $s^0$ in the sum, from this method or any other method?
 A: 
We can apply the binomial series expansion and obtain provided $\left|\frac{D}{C}s\right|<1$
  \begin{align*}
\left(\frac{A+Bs}{C+Ds}\right)^N&=\frac{1}{C^N}\cdot\frac{(A+Bs)^N}{\left(1+\frac{D}{C}s\right)^N}\\
&=\frac{1}{C^N}\cdot(A+Bs)^N\sum_{j=0}^\infty\binom{-N}{j}\left(\frac{D}{C}s\right)^j\tag{1}
\end{align*}
We obtain from (1) the coefficient of $s^0$, denoted with $[s^0]$, which is
  \begin{align*}
[s^0]\left(\frac{A+Bs}{C+Ds}\right)^N=\frac{1}{C^N}\cdot A^N\cdot \binom{-N}{0}=\color{blue}{\left(\frac{A}{C}\right)^N}
\end{align*}
  since we have to select the coefficient of $s^0$ of $(A+Bs)^N$ which is $A^N$ times the summand with $j=0$ which is $\binom{-N}{0}\cdot\left(\frac{D}{C}\right)^0=1$, multiplied with $\frac{1}{C^N}$.

A: This idea suddenly stroke my mind.
We know that
$$\left(\frac{A+Bs}{C+Ds}\right)^N=\sum_{n=-\infty}^{\infty}s^n P_n$$
And since $\forall n \notin \{0,1,\dots,N\}; P_n=0$,
$$\left(\frac{A+Bs}{C+Ds}\right)^N=\sum_{n=0}^{N}s^n P_n$$
So
$$\frac{(A+Bs)^N}{(C+Ds)^N}=\sum_{n=0}^{N}s^n P_n \implies (A+Bs)^N=(C+Ds)^N \sum_{n=0}^{N}s^n P_n$$
$$\sum_{n=0}^N \binom{N}{n} A^{N-n} B^n s^n=\sum_{n=0}^N \binom{N}{n} C^{N-n} D^n s^n \times \sum_{n=0}^{N}s^n P_n $$
If we keep only terms with $s^0$ (which is setting $n=0$), we get
$$A^N=P_0 C^N\implies P_0=(\frac{A}{C})^N$$
