Let $n$ and $a$ be natural numbers. How to prove the following for $x \in [0, 1)$?
$$ (1-x)^{n+a} \sum_{j=0}^\infty \binom{n+j-1}{j}\binom{n+j}{a} x^j = \sum_{j=0}^a \binom{n}{a-j}\binom{a-1}{j} x^j $$
Context: This came up while solving this question.
Numerical evidence on Desmos.
I tried expanding the $(1-x)^{n+a}$ with binomial formula. Or maybe since we're after a sum from $0$ to $a$, expand $(1-x)^{n}$ and $(1-x)^{a}$ separately?
With standard notations (Pochhammer's symbol, Gaussian Hypergeometric Function, $\Gamma$-function and Roy's strategy in https://dx.doi.org/10.2307/2323500 ) the left hand side is essentially $$ \sum_{j\ge 0} \binom{n+j-1}{j}\binom{n+j}{a}x^j = \sum_{j\ge 0} \frac{\Gamma(n+j)\Gamma(n+j+1)}{\Gamma(n)\Gamma(1+a)\Gamma(n+j-a+1)j!} x^j $$ $$ = \frac{1}{\Gamma(1+a)}\sum_{j\ge 0} \frac{(n)_j\Gamma(n+j+1)}{\Gamma(n+j-a+1)j!} x^j $$ $$ = \frac{\Gamma(n+1)}{\Gamma(1+a)}\sum_{j\ge 0} \frac{(n)_j(n+1)_j}{\Gamma(n+j-a+1)} \frac{x^j}{j!} $$ $$ = \frac{\Gamma(n+1)}{\Gamma(1+a)\Gamma(n-a+1)}\sum_{j\ge 0} \frac{(n)_j(n+1)_j}{(n-a+1)_j} \frac{x^j}{j!} $$ $$ = \frac{\Gamma(n+1)}{\Gamma(1+a)\Gamma(n-a+1)}{}_2F_1(n,n+1; n-a+1,x) $$ With a well-known relation for the Gaussian hypergeometric function, eq. B.3 in https://doi.org/10.1016/S0377-0427(00)00267-3 $$ = \frac{\Gamma(n+1)}{\Gamma(1+a)\Gamma(n-a+1)}(1-x)^{-n-a}{}_2F_1(1-a,-a; n-a+1,x) $$ If $a$ is a positive integer, this is a terminating series because "numerators" in the hypergeometric series are negative intgers: $$ = \frac{\Gamma(n+1)}{\Gamma(1+a)\Gamma(n-a+1)}(1-x)^{-n-a}\sum_{j=0}^{a} \frac{(1-a)_j(-a)_j}{(n-a+1)_j}\frac{x^j}{j!} $$ $$ = \frac{\Gamma(n+1)}{\Gamma(1+a)}(1-x)^{-n-a}\sum_{j=0}^{a} \frac{(1-a)_j(-a)_j}{\Gamma(n-a+1+j)}\frac{x^j}{j!} $$ $$ = \frac{\Gamma(n+1)}{\Gamma(1+a)}(1-x)^{-n-a}\sum_{j=0}^{a} (-)^j\frac{\Gamma(a)(-a)_j}{\Gamma(a-j)\Gamma(n-a+1+j)}\frac{x^j}{j!} $$ $$ = \frac{\Gamma(n+1)}{\Gamma(1+a)}(1-x)^{-n-a}\sum_{j=0}^{a} \frac{\Gamma(a)\Gamma(a+1)}{\Gamma(a+1-j)\Gamma(a-j)\Gamma(n-a+1+j)}\frac{x^j}{j!} $$ $$ = \Gamma(n+1)(1-x)^{-n-a}\sum_{j=0}^{a} \frac{\Gamma(a)}{\Gamma(a+1-j)\Gamma(a-j)\Gamma(n-a+1+j)}\frac{x^j}{\Gamma(j+1)} $$ $$ = (1-x)^{-n-a}\sum_{j=0}^{a} \binom{n}{a-j}\binom{a-1}{j}x^j $$ $\square$
We need to show
$$ (1-x)^{n+a} \sum_{m=0}^\infty \binom{n+m-1}{m}\binom{n+m}{a} x^m = \sum_{j=0}^a \binom{n}{a-j}\binom{a-1}{j} x^j. $$
Instead, let's move $(1-x)^{n+a}$ to the RHS using a standard expansion $\frac{1}{(1-x)^{n+1}} = \sum\limits_{i=0}^\infty \binom{n+i}{i} x^i$: $$ \sum_{m=0}^\infty \binom{n+m-1}{m}\binom{n+m}{a} x^m = \sum\limits_{i=0}^\infty \binom{n+a-1+i}{i} x^i\sum_{j=0}^a \binom{n}{a-j}\binom{a-1}{j} x^j. $$
Looking at the coefficient near $x^m$ at both sides, we need to show that $$ \binom{n+m-1}{m} \binom{n+m}{a} = \sum\limits_{j} \binom{n}{a-j} \binom{a-1}{j} \binom{n+a-1+m-j}{n+a-1}. $$
Using $\binom{n}{k} = [x^k](1+x)^n$, the RHS rewrites as $$ \sum\limits_j \binom{a-1}{j}[x^a]x^j(1+x)^n [y^{n+a-1}](1+y)^{n+a-1+m-j} $$ Factoring out $(1+x)^n(1+y)^{n+m}$, this rewrites into $$ [x^a y^{n+a-1}] (1+x)^n (1+y)^{n+m} \sum\limits_j \binom{a-1}{j} x^j (1+y)^{a-1-j}. $$ Using the binomial expansion $\sum\limits_k \binom{n}{k}x^ky^{n-k}= (x+y)^n$, it collapses into $$ [x^a y^{n+a-1}](1+x)^n(1+y)^{m+m}(1+x+y)^{a-1}. $$ Expanding $(1+x+y)^{a-1}$ as $((1+x)+y)^{a-1}$ instead of $((1+y)+x)^{a-1}$, we get $$ [x^a y^{n+a-1}] (1+y)^{n+m}\sum\limits_{j} \binom{a-1}{j} (1+x)^{n+j} y^{a-1-j} = \sum\limits_j \binom{a-1}{j} \binom{n+j}{a} \binom{n+m}{n+j}. $$ In this formula, $\binom{n+m}{n+j}\binom{n+j}{a} = \binom{n+m}{a,n+j-a,m-j} = \binom{n+m}{a}\binom{n+m-a}{m-j}$, thus the sum rewrites into $$ \binom{n+m}{a} \sum\limits_j \binom{a-1}{j} \binom{n+m-a}{m-j} $$
Now, the remaining sum rewrites as $$\begin{align} \sum\limits_j \binom{a-1}{j}\binom{n+m-a}{m-j} &= \sum\limits_j \binom{a-1}{j} [x^{m}] x^j (1+x)^{n+m-a} \\ &= [x^m] (1+x)^{n+m-a} \sum\limits_{j} \binom{a-1}{j} x^j \\ &= [x^m] (1+x)^{n+m-a} (1+x)^{a-1} = \binom{n+m-1}{m}. & \square \end{align}$$
We seek to show that with $n$ and $a$ natural numbers,
$$(1-x)^{n+a} \sum_{j\ge 0} {n+j-1\choose j} {n+j\choose a} x^j = \sum_{j=0}^a {n\choose a-j} {a-1\choose j} x^j.$$
Note that we may assume that $a\ge 1$ because for $a=0$ we get
$$(1-x)^n \frac{1}{(1-x)^n} = 1 = {n\choose 0} {-1\choose 0} x^0.$$
The coefficient on $x^q$ where $q\ge 0$ of the LHS is
$$[x^q] (1-x)^{n+a} \sum_{j\ge 0} {n+j-1\choose j} {n+j\choose a} x^j \\ = \sum_{j\ge 0} {n+j-1\choose j} {n+j\choose a} (-1)^{q-j} {n+a\choose q-j}.$$
Observe carefully that with the third binomial coefficient going zero when $j\ge q$ because $n+a$ is non-negative what we have here is a polynomial in $n$. Therefore it will suffice to evaluate for $n\gt a$ because equality on an infinity of values for two polynomials signifies equality for all arguments. (The coefficient is also a polynomial in $n$ in the proposed closed form.)
Replacing by extractors,
$$[z^a] (1+z)^n [w^q] (1+w)^{n+a} \sum_{j\ge 0} {n+j-1\choose j} (1+z)^j (-1)^{q-j} w^j \\ = (-1)^q [z^a] (1+z)^n [w^q] (1+w)^{n+a} \frac{1}{(1+w(1+z))^n} \\ = (-1)^q [z^a] (1+z)^n [w^q] (1+w)^a \frac{1}{(1+wz/(1+w))^n} \\ = (-1)^q [w^q] (1+w)^a \sum_{p=0}^a {n\choose a-p} {n-1+p\choose p} (-1)^p \frac{w^p}{(1+w)^p} \\ = (-1)^q \sum_{p=0}^a {n\choose a-p} {n-1+p\choose p} (-1)^p {a-p\choose q-p}.$$
Now note that
$${n\choose a-p} {a-p\choose q-p} = \frac{n!}{(n+p-a)! \times (q-p)! \times (a-q)!} = {n\choose a-q} {n-a+q\choose q-p}.$$
We have the first factor. The pair is zero when $q\gt a$ so we will assume $q\le a.$ The remaining sum is
$$(-1)^q \sum_{p=0}^a {n-1+p\choose p} (-1)^p {n-a+q\choose q-p}.$$
With $n-a+q$ a positive number the second binomial coefficient is zero when $q\lt p$ which certainly holds when $p\gt a.$ Hence we may extend $p$ to infinity to get
$$(-1)^q [z^q] (1+z)^{n-a+q} \sum_{p\ge 0} {n-1+p\choose p} (-1)^p z^p \\ = (-1)^q [z^q] (1+z)^{n-a+q} \frac{1}{(1+z)^n} = (-1)^q [z^q] \frac{1}{(1+z)^{a-q}} = {a-1\choose q}.$$
This is the second factor and we may conclude.
