Some hypergeometric transformation

Solution 1:

We represent the binomial identity \begin{align*} \sum_{k\geq 0}&\binom{m+r}{m-n-k}\binom{n+k}{n}x^{m-n-k}y^k\\ &=\sum_{k\geq 0}\binom{-r}{m-n-k}\binom{n+k}{n}(-x)^{m-n-k}(x+y)^k\tag{1} \end{align*} using hypergeometric series by following the presentation in Concrete Mathematics.

LHS

We start with the LHS of (1) and calculate the quotient $\frac{t_{k+1}}{t_{k}}$ of consecutive terms of the sum. We obtain \begin{align*} t_k&=\binom{m+r}{m-n-k}\binom{n+k}{n}x^{m-n-k}y^k\\ &=\frac{(m+r)!(n+k)!}{(m-n-k)!(r+n+k)!n!k!}x^{m-n-k}y^k\\ \\ \frac{t_{k+1}}{t_k}&=\frac{(n+k+1)(m-n-k)}{(r+n+k+1)(k+1)}\,\frac{y}{x}\\ &=\frac{(k+\color{blue}{n+1})(k+\color{blue}{n-m})}{(k+\color{blue}{n+r+1})(k+1)}\left(\color{blue}{-\frac{y}{x}}\right) \end{align*} The first summand $k=0$ of the LHS of (1) is $\binom{m+r}{m-n}x^{m-n}$ and we conclude

\begin{align*} \sum_{k\geq 0}&\binom{m+r}{m-n-k}\binom{n+k}{n}x^{m-n-k}y^k\\ &=\binom{m+r}{m-n}x^{m-n}\color{blue}{F\left(n+1,n-m;n+r+1;-\frac{y}{x}\right)}\tag{2.1} \end{align*}

RHS

We proceed in the same way with the RHS and obtain \begin{align*} u_k&=\binom{-r}{m-n-k}\binom{n+k}{n}(-x)^{m-n-k}(x+y)^k\\ &=\frac{(-r)!(n+k)!}{(m-n-k)!(-r-m+n+k)!n!k!}\left(-x\right)^{m-n-k}(x+y)^k\\ \\ \frac{u_{k+1}}{u_k}&=\frac{(n+k+1)(m-n-k)(-1)}{(-r-m+n+k+1)(k+1)}\,\frac{x+y}{x}\\ &=\frac{(k+\color{blue}{n+1})(k+\color{blue}{n-m})}{(k+\color{blue}{n-m-r+1})(k+1)}\,\color{blue}{\frac{x+y}{x}} \end{align*} The first summand $k=0$ of the RHS of (1) is $\binom{-r}{m-n}(-x)^{m-n}$ and we conclude

\begin{align*} \sum_{k\geq 0}&\binom{-r}{m-n-k}\binom{n+k}{n}(-x)^{m-n-k}(x+y)^{k}\\ &=\binom{-r}{m-n}(-x)^{m-n}\color{blue}{F\left(n+1,n-m;n-m-r+1;\frac{x+y}{x}\right)}\tag{2.2} \end{align*}

Replacing $n$ with $q$ in (2.1) and (2.2) we obtain from (1) \begin{align*} &\color{blue}{F\left(q+1,q-m;q+r+1;-\frac{y}{x}\right)=\binom{-r}{m-q}\binom{m+r}{m-q}^{-1}(-1)^{m-q}}\\ &\qquad \color{blue}{\cdot F\left(q+1,q-m;q-m-r+1;\frac{x+y}{x}\right)}\tag{2.3} \end{align*}

Substitutions

Finally we show the identity (2.3) can be written as \begin{align*} \color{blue}{F(a,-n;c;z)=\frac{(a-c)^{\underline n}}{(-c)^{\underline n}}F(a,-n;1-n+a-c;1-z)}\tag{3.1} \end{align*} Comparing the LHS of (2.3) and (3.1) we use the substitutions \begin{align*} a&:=q+1\\ n&:=-q+m\\ c&:=q+r+1\tag{3.2}\\ z&:=-\frac{y}{x} \end{align*}

With these substitutions the LHS of (2.3) and (3.1) coincide. The upper parameters $q+1$ and $q-m$ are the same at both sides of (2.3). We now use the substitutions to derive the argument, the lower parameter and the factor in front of the hypergeometric series.

• From $z=-\frac{y}{x}$ we see \begin{align*} 1-z=1+\frac{y}{x}=\frac{x+y}{x} \end{align*} and the arguments coincide.

• Looking at the lower parameter $n-m-r+1$ in (2.3) we obtain with (3.2) \begin{align*} q-m-r+1&=(-n)-r+1\\ &=(-n)-(c-q-1)+1\\ &=(-n)-(c-a)+1\\ &=1-n+a-c \end{align*} in accordance with the lower parameter of the RHS in (3.1).

• We obtain \begin{align*} \binom{-r}{m-q}&=\binom{a-c}{n}=(a-c)^{\underline{n}}\\ \binom{m+r}{m-q}(-1)^{m-q}&=\binom{-q-r-1}{m-q}(-1)^{m-q}=\binom{-c}{n}\\ &=(-c)^{\underline{n}} \end{align*}

and the claim (3.1) follows.