Prove that for every real $x$ and positive integer $n$ the following is true: $$[x]+\left[x+\frac{1}{n}\right]+\cdots+\left[x+\frac{n-1}{n}\right]=[nx].$$
Here, if $y$ is any real, then $[y]$ denotes the integer part of $y$ (that is, the greatest integer $m$ such that $m \leq y$).
It's called Hermite's identity. Let $$f(x):=[x]+\left[x+\frac{1}{n}\right]+\cdots+\left[x+\frac{n-1}{n}\right]-[nx].$$ It's easy to check that: $$f\left(x+\frac{1}{n}\right)=\left[x+\frac{1}{n}\right]+\cdots+\left[x+\frac{n-1}{n}\right]+[x+1]-\left[n(x+\frac{1}{n})\right]=f(x).$$ Here I used that $[x+k]=[x]+k$ for every $k\in \mathbb{Z}$ .
So our function $f$ is periodic with period $1/n$. It is enough to explore $f$ on interval $[0,1/n)$. But if $x\in[0,1/n)$ then $f(x)\equiv 0$ because every term is equal to zero. Q.E.D.