How to calculate this sum with binomials? In this paper An  Improved Algorithm  for Decentralized Extrema-Finding in  Circular Configurations of  Processes from 1979, in chapter with analyze of average number of passes messages I found this equation:
$$\sum_{i=k}^{n-1}k \frac{{i-1}\choose{k-1}}{{n-1}\choose{k-1}} \times \frac{n-i}{n-k} = \frac{n}{k+1}$$
WolframAlpha solution.
Can anyone explain me why it's equal and how can I calculate it?
In paper they write only that this left side "can be simplified to" right side.
Edit 1:
I think that main problem is to understand what this sum is:
$$\sum_{i=k}^{n-1} {{i-1}\choose{k-1}} \times (n-i)$$
WolframAlpha.
Because:
$$\sum_{i=k}^{n-1}k \frac{{i-1}\choose{k-1}}{{n-1}\choose{k-1}} \times \frac{n-i}{n-k} = \frac{k}{{{n-1}\choose{k-1}}(n-k)} \sum_{i=k}^{n-1} {{i-1}\choose{k-1}} \times (n-i)$$
And according to WolframAlpha:
$$\sum_{i=k}^{n-1} {{i-1}\choose{k-1}} \times (n-i) = \frac{n(n-k){{n-1}\choose{k-1}}}{k(k+1)} $$
So everything could be nicely reduced.
 A: Have you seen this formula, sometimes known as the hockey-stick formula
$$\sum_{i=k}^{n-1}{i-1\choose k-1}={n-1\choose k}$$
It is proved by identifying the binomial coefficients in Pascal's triangle, noting that ${k-1\choose k-1}={k\choose k}$, and cascading down the diagonal.  
Another relevant formula is $i{i-1 \choose k-1} = k{i\choose k}$.  Put them together and you get 
$$(n-i){i-1\choose k-1} = n{i-1\choose k-1} - k{i\choose k}$$
then sum them up separately.
A: Not a complete solution
\begin{align}
\sum_{i=k}^{n-1}k \frac{{i-1}\choose{k-1}}{{n-1}\choose{k-1}} \times \frac{n-i}{n-k} &= \sum_{i=k}^{n-1} k\frac{(i-1)!(k-1)!(n-k)!~}{(k-1)!(i-k)!(n-1)!} \times \frac{n-i}{n-k}\\
&= \sum_{i=k}^{n-1} k\frac{(i-1)!(n-k-1)!(n-i)}{(i-k)!(n-1)!} \\
&= k \cdot \frac{(n-k-1)!}{(n-1)!} \cdot\sum_{i=k}^{n-1} \frac{(i-1)!(n-i)}{(i-k)!} \\
\end{align}
I'm not sure where to proceed from here, either. 
A: $$\begin{align}
\sum_{i=k}^{n-1}\color{orange}{\binom {i-1}{k-1}(n-i)}
&=\sum_{i=k}^{n-1}\sum_{j=i}^{n-i}\binom{i-1}{k-1}\\
&=\sum_{j=k}^{n-1}\sum_{i=k}^j\binom {i-1}{k-1}\\
&=\sum_{j=k}^{n-1}\binom jk\\
&=\binom n{k+1}\\
&=\frac 1{k+1}\binom {n}{k}\frac{n-k}1\\
&=\frac {\color{red}n}{\color{red}{(k+1)}\color{blue}k}\color{green}{\binom {n-1}{k-1}}\frac{\color{green}{n-k}}1\\
\sum_{i=k}^{n-i}\color{blue}k\frac {\displaystyle\color{orange}{\binom{i-1}{k-1}}}{\displaystyle\color{green}{\binom {n-1}{k-1}}}\cdot \frac {\color{orange}{n-i}}{\color{green}{n-k}}&=
\color{red}{\frac n{k+1}}\qquad\color{red}\blacksquare
\end{align}$$
