I am looking to calculate
$$\sum^{n}_{j = 0}{{a + j}\choose{b}}$$
where $a, b \in \mathbb{Z}$. I realize that if $b = a$ then this is proven here with a different labeling of variables, but in my case $a$ and $b$ are different.
Among the things I have tried that did not work were the Snake Oil method and applying Hockey-Stick Identities (such as this one).
Either an analytic or combinatorial proof is fine.
Hints:
Notice that if $a<b=a+c$ then it is the same thing as the case $a=b$ because
$$\sum _{j=0}^n\binom{a+j}{b}=\sum _{j=-c}^{n-c}\binom{a+c+i}{b}=\sum _{j=0}^{n-c}\binom{b+i}{b},$$
can you conclude?
For the case in which $b<a$ then you can write the sum as suggested by peterwhy in the comments i.e.,
$$\sum _{j=0}^{a-1}\binom{j}{b}+\sum _{j=0}^{n}\binom{a+j}{b}=\sum _{j=0}^{a+n}\binom{j}{b},$$ can you conclude?
This is not the most direct approach but it does show where hypergeometric functions appear as mentioned by @DrSonnhardGraubner.
We write $$ \begin{align} \sum_{j=0}^n\binom{a+j}{b} &=\sum_{j=0}^n\frac{\Gamma(a+1+j)}{\Gamma(b+1)\Gamma(a-b+1+j)}\\ &=\frac{\Gamma(a+1)}{\Gamma(b+1)\Gamma(a-b+1)}\sum_{j=0}^n\frac{(a+1)_j}{(a-b+1)_j}\\ &=\frac{\Gamma(a+1)}{\Gamma(b+1)\Gamma(a-b+1)}\sum_{j=0}^n\frac{(a+1)_j(1)_j}{(a-b+1)_j\,j!}, \end{align} $$ where $(s)_n=\Gamma(s+n)/\Gamma(s)$ is the Pochhammer symbol and $(1)_n=n!$. The result is a partial hypergeometric series, which may be evaluated using DLMF 16.2.4 to give $$ \sum_{j=0}^n\binom{a+j}{b}=\frac{\Gamma(a+1)}{\Gamma(b+1)\Gamma(a-b+1)}\frac{(a+1)_n(1)_n}{(a-b+1)_n\,n!}{_3F_2}\left({-n,1,b-a-n\atop -n,-a-n};1\right). $$ After some simplification $$ \sum_{j=0}^n\binom{a+j}{b}=\frac{\Gamma(a+1+n)}{\Gamma(b+1)\Gamma(a-b+1+n)}{_2F_1}\left({1,b-a-n\atop -a-n};1\right). $$ Gauss's hypergeometric theorem then permits us to write $$ \sum_{j=0}^n\binom{a+j}{b}=\frac{\Gamma(a+1+n)}{\Gamma(b+1)\Gamma(a-b+1+n)}\frac{\Gamma(-a-n)\Gamma(-1-b)}{\Gamma(-a-n-1)\Gamma(b)}, $$ which can be further reduced to give a sum of two binomial coefficients.
