Sum of binomial terms, but with fixed exponents and a variable proportion I am working on a problem where I have to calculate the summation of "binomial terms":
$\sum_{t=0}^{N} (t/N)^k (1 - t/N)^{N-k}$
where k and N are fixed (we are indexing over t). Equivalently, we can think of this as
$\sum_{t=0}^{1} (t)^k (1 - t)^{N-k}$ where $t=0, \, \frac{1}{N}, \, \frac{2}{N}, ..., 1$
In plain English, we are summing binomial terms over equally-spaced proportions, e.g. for N=10 we would be summing over t = 0/10, 1/10, 2/10, ..., 9/10, 10/10 (for some known k, N).
Note that the usual binomial expansion handles the summation for indexing by variable k, but in my case k will be fixed.
For context, I am trying to determine the average of a discretely sampled Bezier curve (sampled across consistent intervals). This requires calculating this sum over all k<N, for which there may even be a shortcut that I'm unaware of. The continuous case is actually easy (just average the control points) but I'm struggling with the discrete case.
Thanks in advance.

I'm not sure there is an exact solution, but I think I found an approximation: If you rewrite the summand using $x=e^{\log x}$ you get
\begin{align}
S_n &= \sum_{t=1}^{n}e^{k \log \frac{t}{n}}e^{(n-k)\log (1-\frac{t}{n})} \approx \sum_{t=1}^{n}e^{k \log \frac{t}{n}}e^{-(n-k)\frac{t}{n}} \\
&= \sum_{k=1}^{n}e^{k \log t-t+\frac{kt}{n} - k \log n} = e^{-k \log n}\sum_{k=1}^{n}t^k e^{-c_1 t}
\end{align}
For some constant $c_1$. I don't know if the last expression exists in the closed form, but it is comparable to the lower incomplete Gamma function:
$$
S_n \approx e^{-k \log n}\int_{0}^{n}x^ke^{-c_1 x}dx = \frac{e^{-k  \log n}}{c_1^k}\int_{0}^{c_1n}v^ke^{-v}dv = \frac{e^{-k  \log n}}{c_1^k}\gamma(k+1, c_1n)
$$
