If $xy = ax + by$, prove the following: $x^ny^n = \sum_{k=1}^{n} {{2n-1-k} \choose {n-1}}(a^nb^{n-k}x^k + a^{n-k}b^ny^k),n>0$ If $xy = ax + by$, prove the following: $$x^ny^n = \sum_{k=1}^{n} {{2n-1-k} \choose {n-1}}(a^nb^{n-k}x^k + a^{n-k}b^ny^k) = S_n$$  for all $n>0$
We'll use induction on $n$ to prove this.

My approach is to use this formula: $$ \frac{k}{r} {{r} \choose {k}} = {{r-1} \choose {k-1}}$$
I'd like to show: $$S_{n} = xy.S_{n-1}$$. Or:
$$\sum_{k=1}^{n} {{2n-1-k} \choose {n-1}}(a^{n}b^{n-k}x^k + a^{n-k}b^{n}y^k) = (ax+by)\sum_{k=1}^{n-1} {{2n-3-k} \choose {n-2}}(a^{n-1}b^{n-k-1}x^k + a^{n-1-k}b^{n-1}y^k)$$
We have:
$$ (ax+by)\sum_{k=1}^{n-1} {{2n-3-k} \choose {n-2}}(a^{n-1}b^{n-k-1}x^k + a^{n-1-k}b^{n-1}y^k)
= \sum_{k=1}^{n-1} {{2n-3-k} \choose {n-2}}(a^{n}b^{n-k-1}x^{k+1} + a^{n-k}b^{n-1}xy^k + a^{n-1}b^{n-k}x^ky + a^{n-1-k}b^ny^{k+1})
= \sum_{k=2}^{n} {{2n-2-k} \choose {n-2}}(\pmb{a^{n}b^{n-k}x^{k}} + a^{n-k+1}b^{n-1}xy^{k-1} + a^{n-1}b^{n-k+1}x^{k-1}y + \pmb{a^{n-k}b^ny^{k}})
= \sum_{k=2}^{n} \frac{n-1}{2n-1-k} {{2n-1-k} \choose {n-1}} [...]
$$
Now we can almost extract the intended term($S^{'}_{n}$):
$$ 
\sum_{k=2}^{n} {{2n-1-k} \choose {n-1}} (a^{n}b^{n-k}x^{k} + a^{n-k}b^ny^{k}) + \sum_{k=2}^{n} {{2n-1-k} \choose {n-1}} (a^{n-k+1}b^{n-1}xy^{k-1} + a^{n-1}b^{n-k+1}x^{k-1}y) + \sum_{k=2}^{n} (\frac{n-1}{2n-1-k}-1) {{2n-1-k} \choose {n-1}} [...]
$$
There is further derivation but not seems very promising.
The idea of this theorem is really interesting. I'm asking for a simpler approach or how i should countinue my proof. Thank you in advance!
 A: The approach by induction is fine. We just need one additional twist (one more induction).
We want to show for $n\geq 1$
\begin{align*}
\color{blue}{(xy)^n=\sum_{k=1}^n\binom{2n-1-k}{n-1}\left(a^nb^{n-k}x^k+a^{n-k}b^ny^k\right)}\tag{1}
\end{align*}
provided
\begin{align*}
\color{blue}{xy=ax+by}
\end{align*}
The base step $n=1$ is easily shown.
Induction hypothesis: $n=N-1$:
We assume the claim (1) is valid for $n=N-1$, i.e. we have
\begin{align*}
(xy)^{N-1}=\sum_{k=1}^{N-1}\binom{2N-3-k}{N-2}\left(a^{N-1}b^{N-1-k}x^k+a^{N-1-k}b^{N-1}y^k\right)\tag{2}
\end{align*}
Induction Step: $n=N$:
Now we want the show the validity of (1) for $n=N$. We consider $(xy)^N=(xy)^{N-1}(xy)$. Multiplication of the right hand side of (2) with $xy$ gives us terms $Ax^ky$ and $Bxy^k$ with constants $A$ and $B$. We obtain
\begin{align*}
xy&=ax+by\\
x^2y&=ax^2+abx+b^2y\\
x^3y&=ax^3+abx^2+ab^2x+b^3y\\
&\vdots\\
x^ky&=\sum_{j=1}^k ab^{k-j}x^j+b^ky\qquad\qquad xy^k=a^kx+\sum_{j=1}^ka^{k-j}by^j\tag{3}
\end{align*}
which can be easily shown by induction.

Putting (3) in (2) we obtain
\begin{align*}
\color{blue}{(xy)^n}&=\left(xy\right)^{n-1}(xy)\\
&=\sum_{k=1}^{N-1}\binom{2N-3-k}{N-2}\left(a^{N-1}b^{N-1-k}x^k+a^{N-1-k}b^{N-1}y^k\right)(xy)\\
&=\sum_{k=2}^{N}\binom{2N-2-k}{N-2}\left(a^{N-1}b^{N-k}x^ky+a^{N-k}b^{N-1}xy^k\right)\tag{4}\\
&=\sum_{k=2}^{N}\binom{2N-2-k}{N-2}\left(a^{N-1}b^{N-k}\left(\sum_{j=1}^kab^{k-j}x^j+b^ky\right)\right.\\
&\qquad\qquad\qquad\qquad\qquad\left.+a^{N-k}b^{N-1}\left(a^kx+\sum_{j=1}^ka^{k-j}by^j\right)\right)\tag{5}\\
&\,\,\color{blue}{=\sum_{k=2}^{N}\binom{2N-2-k}{N-2}\left(\sum_{j=1}^ka^Nb^{N-j}x^j+a^{N-1}b^Ny\right.}\\
&\qquad\qquad\qquad\qquad\qquad\qquad\color{blue}{\left.+a^Nb^{N-1}x+\sum_{j=1}^ka^{N-j}b^Ny^j\right)}\tag{6}
\end{align*}

Comment:

*

*In (4) we shift the index and start with $k=2$.


*In (5) we use the identities from (3).

In order to show that (1) and (6) are equal we show equality of coefficients of the polynomials in $x$ and $y$. We denote with $[x^t]$ the coefficient of $x^t$ and obtain from (1) for $1\leq t\leq N$:
\begin{align*}
[x^t](xy)^N=\binom{2N-1-t}{N-1}a^Nb^{N-t}\tag{7}
\end{align*}
We obtain from (6)
\begin{align*}
[x^t](xy)^N&=\sum_{k=t}^N\binom{2N-2-k}{N-2}a^Nb^{N-t}\\
&\qquad+\sum_{k=2}^N\binom{2N-2-k}{N-2}a^Nb^{N-1}[[t=1]]\tag{8}
\end{align*}
with $[[t=1]]$ denoting Iverson brackets which is $1$ iff $t=1$ and zero otherwise.

Equating (7) and (8) we finally have to show that for $1\leq t\leq N$ we have
\begin{align*}
\color{blue}{\sum_{k=t}^N\binom{2N-2-k}{N-2}+\sum_{k=2}^N\binom{2N-2-k}{N-2}[[t=1]]
=\binom{2N-1-t}{N-1}}\tag{9}
\end{align*}

We obtain for $t\geq 2$:
\begin{align*}
\color{blue}{\sum_{k=t}^N}&\color{blue}{\binom{2N-2-k}{N-2}}\\
&=\sum_{k=0}^{N-t}\binom{2N-2-t-k}{N-2}\\
&=\sum_{k=0}^{N-t}\binom{N-2+k}{N-2}\tag{$k\to N-t-k$}\\
&=\sum_{k=0}^{N-t}[z^{N-2}](1+z)^{N-2+k}\\
&=[z^{N-2}](1+z)^{N-2}\sum_{k=0}^{N-t}(1+z)^{k}\\
&=[z^{N-2}](1+z)^{N-2}\frac{(1+z)^{N-t+1}-1}{(1+z)-1}\\
&=[z^{N-1}]\left((1+z)^{2N-t-1}-(1+z)^{N-2}\right)\\
&=[z^{N-1}](1+z)^{2N-t-1}\\
&\,\,\color{blue}{=\binom{2N-t-1}{N-1}}
\end{align*}
showing the claim (9) is valid for $t\geq 2$.


Since the above result is also valid for $t=1$ we use it to show the case $t=1$:
\begin{align*}
\color{blue}{2\sum_{k=2}^N}&\color{blue}{\binom{2N-2-k}{N-2}}\\
&=2\sum_{k=1}^N\binom{2N-2-k}{N-2}-2\binom{2N-3}{N-2}\\
&=2\binom{2N-2}{N-1}-2\,\frac{N-1}{2N-2}\binom{2N-2}{N-1}\\
&\,\,\color{blue}{=\binom{2N-2}{N-1}}
\end{align*}
showing the claim (9) is valid for $t=1$.

Showing the validity of $[y^t](xy)^N$, $1\leq t\leq N$ follows the same lines.
