Simplifying expression using summation

Question

Is there a way can simplify the following expression further?

$$ \sum_{i=k}^n {i \choose k} a^i$$

where $a$ is some positive real number.

I am aware that

$$ \sum_{i=k}^n {i \choose k} = {n+1 \choose k+1}$$

but the $a^i$ is throwing me off

Answer 1

$$\sum_{i=k}^n {i \choose k} a^i=\frac{a^k}{(1-a)^{k+1}}-a^{n+1} \binom{n+1}{k} \,\, _2F_1(1,n+2;n-k+2;a)$$ where appears the gaussian hypergeometric function (which is far away from elementary functions).

Answer 2

If you replace the upper index with $\infty$, the resulting sum is $$\sum_{i=k}^\infty {i \choose k} a^i=\frac{a^k}{(1-a)^{k+1}},$$ but I don't think the $n$th partial sum simplifies to anything nice.