TITLE: Formula for finite sum $\sum_{1\leq\alpha_{1}<\alpha_{2}<\ldots<\alpha_{k}\leq n}\frac{\alpha_{1}+\alpha_{2}+\ldots+\alpha_{k}}{k}$
QUESTION [10 upvotes]: Let $n$ and $k$ be positive integers, $k\leq n$. Is there a formula for this sum?
$$
\sum_{(\alpha_{1}, \alpha_{2},\ldots, \alpha_{k}):\,\, 1\leq\alpha_{1}<\alpha_{2}<\ldots<\alpha_{k}\leq n}\frac{\alpha_{1}+\alpha_{2}+\ldots+\alpha_{k}}{k}
$$
Thank you very much in advance!

REPLY [8 votes]: Let $A$ be the collection of $k$-tuple $1 \le \alpha_1 < \alpha_2 < \cdots < \alpha_k \le n$.
Since there is a bijection on $A \ni (\alpha_i) \mapsto (n + 1 - \alpha_i) \in A$ which send 
the sum $\frac{\sum_{i} \alpha_i}{k}$ to $n+1 - \frac{\sum_{i} \alpha_i}{k}$, we have:
$$\sum_{\alpha \in A} \frac{\alpha_1 + \alpha_2 + \cdots + \alpha_k}{k} = \sum_{\alpha \in A} \frac{n+1}{2} = \binom{n}{k}\frac{n+1}{2}$$