Sum of a floor function

Question

I have the following function :

$$ f(x)= \sum_{i=0}^{n-1}\left\lfloor {\frac{x+i}{n}}\right\rfloor $$

I have to show that :

$\forall_{x}\in \mathbb{R}, f(x+1)=f(x)+1$

I proved that : $\forall x\in [0,1[$, $f(x)=0$, but I struggle finding something to do with it.

Thank you for any help.

Answer 1

Notice that you can write $f(x+1)$ to look like something involving $f(x)$ by reindexing the sum and breaking it apart:$$f(x+1) = \sum_{i=0}^{n-1} \left\lfloor \frac{(x+1)+i}{n} \right\rfloor = \sum_{i=1}^{n} \left\lfloor \frac{x+i}{n} \right\rfloor = \left(\sum_{i=0}^{n-1} \left\lfloor \frac{x+i}{n} \right\rfloor \right) + \left\lfloor \frac{x+n}{n} \right\rfloor - \left\lfloor \frac{x+0}{n} \right\rfloor = f(x)+ \left\lfloor \frac{x}{n}+1 \right\rfloor - \left\lfloor \frac{x}{n} \right\rfloor = f(x)+1$$