I was working on my probability theory homework and I found probability density function that looks as following
$$\sum_{t=k}^{n} \binom{n}{t} \cdot a^{t-1} \cdot (1 - a)^{n - t - 1} \cdot (t - n \cdot a)$$ and I was wondering if it can be simplified somehow. Wolfram said "yes" and gave me this answer
$$\sum_{t=k}^n a^{t - 1} (1 - a)^{n - t - 1} \cdot (t - a \cdot n) \cdot \binom{n}{t} = k \cdot a^{k - 1} (1 - a)^{n - k} \cdot \binom{n} {k}$$
For shortness, instead of writing $F(x)$ I wrote $a$ (its nature is really not important here, it is just distribution function for some other random variable). I tried integrating this thing on the right and was hoping to get distribution function and it would at least prove its correctness. However, all I was able to get is just some recurrent formula for integral that did not help. But is there any simple and pretty way to simplify it? I would love to hear your suggestions, thanks!
Here's the second part: \begin{align} \int_0^1 k a^{k-1} (1-a)^{n-k} \binom{n}{k} \mathrm{d}a &= k \binom{n}{k} \int_0^1 a^{k-1} (1-a)^{n-k} \mathrm{d}a \\ &= k \binom{n}{k} B(k,n-k+1) \\ &= k \binom{n}{k} \frac{\Gamma(k)\Gamma(n-k+1)}{\Gamma(n+1)} \\ &= k \binom{n}{k} \frac{(k-1)!(n-k)!}{n!} \\ &= 1 \end{align}