# Formula for finite sum $\sum_{1\leq\alpha_{1}<\alpha_{2}<\ldots<\alpha_{k}\leq n}\frac{\alpha_{1}+\alpha_{2}+\ldots+\alpha_{k}}{k}$

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!

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:

$$\sum_{\alpha \in A} \frac{\alpha_1 + \alpha_2 + \cdots + \alpha_k}{k} = \sum_{\alpha \in A} \frac{n+1}{2} = \binom{n}{k}\frac{n+1}{2}$$

• For displayed equations, you need to place the punctuation inside the double dollar signs (preferably spaced apart by \;), otherwise it's placed on the next line as in this case. Feb 20, 2013 at 17:45
• @joriki thanks. Feb 20, 2013 at 17:50

Since $k$ is fixed, we can pull it out of the sum. Then it's just a matter of adding up the elements of all $k$-combinations of the natural numbers up to $n$. There are $\binom nk$ such combinations, containing $k$ numbers each, and each number from $1$ to $n$ appears the same number of times, so we can replace them all by their average $(n+1)/2$, so the total sum is $\binom nk\frac{n+1}2$.

• This is correct, but achille's answer states the fact that I was going to mention that each choice of $(\alpha_k)$ is matched with another $(n+1-\alpha_k)$. (+1)
– robjohn
Feb 20, 2013 at 17:52