[Calculation 4] Saalschütz’s Theorem

 
Proof. We recall the Euler Transformation Formula:
$$(1-z)^{a+b-c} {}_2 F_1 (a,b;c;z) = {}_2 F_1 (c-a,c-b;c;z).$$
Equating the coefficients of $z^n$ from both sides, we get
\begin{equation}\tag{1}
\sum_{k=0}^n \frac{(a)_k (b)_k}{(c)_k k!} \frac{(c-a-b)_{n-k}}{(n-k)!}
=
\frac{(c-a)_n (c-b)_n}{(c)_n n!}
\end{equation}
where we used the binomial theorem
$$
(1-z)^{a+b-c} = \sum_{k=0}^\infty \frac{(c-a-b)_k}{k!}z^k.
$$
In view of the relations
$$
\frac{n!}{(n-k)!} = (-1)^k (-n)_k
$$
and
$$
(c-a-b)_{n-k} = \frac{(c-a-b)_n}{(-1)^k (1+a+b-c-n)_k},
$$
we can rewrite (1) as
$$
\sum_{k=0}^n \frac{(a)_k (b)_k}{(c)_k k!} \frac{(c-a-b)_n (-n)_k}{(1+a+b-c-n)_k n!}
=
\frac{(c-a)_n (c-b)_n}{(c)_n n!}.
$$
Therefore we finally obtain
\begin{eqnarray*}
{}_3 F_2 (a,b,-n;c,1+a+b-c-n;1)
&=&
\sum_{k=0}^\infty \frac{(a)_k (b)_k (-n)_k}{(c)_k (1+a+b-c-n)_k k!}\\
&=&
\sum_{k=0}^n \frac{(a)_k (b)_k (-n)_k}{(c)_k (1+a+b-c-n)_k k!}\\
&=&
\frac{(c-a)_n (c-b)_n}{(c)_n (c-a-b)_n}.
\end{eqnarray*}
This proves the theorem.$\square$
 
References.
[1] W. N. Bailey, Generalized Hypergeometric Series, Cambridge University Press, 1964, p.9.
[2] Leun Kim, Euler’s Transformation Formula