What is a closed form for this combinatorial sum?

Question

Primarily I wanted to solve the recurrence system $$c_k=\frac 1p{p\choose k}-c_{k-1},\\c_0=0,$$ for any integer $$1\le k\le \frac{p-1}{2},$$ where $p$ is any odd prime. I was only able to see that $c_k$ could be represented by the sum $$\sum_{i=0}^{k-1}\frac{(-1)^i}{p}{p\choose {k-i}}.$$

However, I would like a closed form for this sum, in terms of elementary functions, if it exists. (Or if you can proceed straight from the recurrence system to a closed form solution, all would still be fine. All I need is the solution in closed form.)

Thank you.

Answer 1

Multiplying both sides of the recurrence by $x^k$ and summing for $k\geq 1$ gives $$ C(x) = \frac{1}{p}((1 + x)^p - 1) + xC(x), $$ i.e, $$ C(x) = \frac{{(1 + x)^p - 1}}{{p(1 + x)}} = \frac{1}{p}\left( {(1 + x)^{p - 1} - \frac{1}{{1 + x}}} \right), $$ where $C(x)$ is the generating function of $c_k$. From this, $$ c_k = \frac{1}{p}\left( {\binom{p - 1}{k} - ( - 1)^k } \right). $$

Answer 2

WolframAlpha claims this sum reduces to $$\frac{(p - k) \binom{p}{k} - (-1)^k p}{p^2},$$ which further simplifies to $$\frac{p \binom{p-1}{k} - (-1)^k p}{p^2}=\frac{\binom{p-1}{k} - (-1)^k}{p}.$$