2. Let $n \in \mathbb{N}^{*}$ be fixed. Solve the equation: $\left\{x+\frac{1}{2 n}\right\}+\left\{x+\frac{3}{2 n}\right\}+\ldots+\left\{x+\frac{2 n-1}{2 n}\right\}=\frac{n(2 x-1)}{2}$.

Supliment G.M.12/2015
Solution. The equation is equivalent to: $n x+\frac{1+3+\ldots+(2 n-1)}{2 n}-\left(\left[x+\frac{1}{2 n}\right]+\left[x+\frac{3}{2 n}\right]+\ldots+\left[x+\frac{2 n-1}{2 n}\right]\right)=n x-\frac{n}{2} \Leftrightarrow$ $\Leftrightarrow\left[x+\frac{1}{2 n}\right]+\left[x+\frac{3}{2 n}\right]+\ldots+\left[x+\frac{2 n-1}{2 n}\right]=n$. If we denote $f(x)=\left[x+\frac{1}{2 n}\right]+\left[x+\frac{3}{2 n}\right]+\ldots+\left[x+\frac{2 n-1}{2 n}\right]$, then $f\left(x+\frac{1}{n}\right)=f(x)+1, \forall x \in \mathbb{R}$ and $f(x)=n, \forall x \in\left[\frac{2 n-1}{2 n}, \frac{2 n+1}{2 n}\right)$, so $f$ is constant on each interval of length $\frac{1}{n}$ of the form $I_{n}=\left[\frac{2 n-1}{2 n}, \frac{2 n+1}{2 n}\right), n \geq 1$. Since $f$ is an increasing function, it follows that the set of solutions of the equation is $I_{n}$.

Scoring.

| Deduce $\left[x+\frac{1}{2 n}\right]+\left[x+\frac{3}{2 n}\right]+\ldots+\left[x+\frac{2 n-1}{2 n}\right]=n$ | $3 \text{p}$ |
| :--- | :--- |
| Solve and find the set of solutions | $4 \text{p}$ |