Let $X_1,...,X_n$ be i.i.d. random variable with a uniform distribution on [0,1]. Denote by $X_{(1)}\leq X_{(2)} \leq \ldots \leq X_{(n)}$ their order statistics.

Given $k\geq 1$ and $u\in[0,k]$, I want a simple formula for $$ p_k(u):=\mathbb{P}[X_{(1)}+\ldots + X_{(k)}\leq u], $$ or at least a simple lower bound on $p_k(u)$. (By simple, I mean that I don't want a k-dimensional integral...)

**Note:**
Of course, the case $k=1$ is easy:
$$p_1(u) = 1-(1-u)^n.$$
And, by using the representation
$$U_{(1)},U_{(2)} = 1-Y^{1/n},1-Y^{1/n}Z^{1/(n-1)},$$
where $X,Y$ are iid $\sim U([0,1])$,
I was able to compute $p_2(u)$ (with the help of Maple):
$$p_2(u) = 1-2(1-\frac{u}{2})^n+ \big(\max(1-u,0)\big)^n$$

Mathematicaworking is incorrect: this is because it is incorrect to treat your $x$ as the 1st order statistic and your $y$ as the second order statistic ... you need the JOINT pdf of the 1st and 2nd order statistics. The OP's result seems correct to me, and very neatly stated too. $\endgroup$ – wolfies Oct 29 '17 at 5:06