Let $a_1,a_2 \in \mathbb{R}$, and let $n,k$ be positive integers. Is there a closed-form expression for $\sum_{j=0}^k \binom{n+j-1}{j} a_1^j a_2^{k-j}$?
Context: This quantity looks very similar to that appearing in the Binomial theorem: $\sum_{j=0}^k \binom{k}{j} a_1^j a_2^{k-j}=(a_1+a_2)^k.$ I have encountered this quantity in the following combinatorial problem:
Let $\mu_1=a_2, \mu_2=\dots=\mu_{n+1}=a_1$. Then the desired quantity is equal to the sum of the elements of the following multiset: $\{\mu_{i_1}\cdots \mu_{i_k} : i_1 \leq i_2 \leq \dots \leq i_k \in [n]\}$.
Best Answer
$$\sum_{j=0}^{k}\binom{n+j-1}{j}\ a^jb^{k-j}=\frac{b^{n+k}}{(a-b)^n}-\frac{1}{b}\binom{n+k}{k+1}\ a^{k+1}\ _2F_1(1,k+1+n;\,k+2;\,\frac {a}{b})$$
$_2F_1$ is a generalized hypergeometric function.