sum of integral parts of real number and fraction

Question

For any real x and positive integer n ,show that

[x] + [x +1/n] + [x + 2/n] + .... + [x + n-1/n] = [nx]

I have used the fact that x-1 < [x] <= x,for all terms and added,but not able to get tight upper bound with it

Answer 1

Let $x=a+t$, where $a$ is an integer, and $0\le t\lt 1$. Suppose that $\frac{k-1}{n}\le t\lt \frac{k}{n}$, where $1\le k\le n$. Then $\lfloor nx\rfloor=na+n-k$.

Now compute the left-hand side. The terms $\left\lfloor a+t+\frac{i}{n}\right\rfloor$ are equal to $a$ if $0\le i\lt k$, and are equal to $a+1$ for $k\le i\le n-1$.

So there are $k$ values of $i$ for which $\left\lfloor a+t+\frac{i}{n}\right\rfloor=a$ and $n-k$ values of $i$ for which $\left\lfloor a+t+\frac{i}{n}\right\rfloor=a+1$. It follows that the left-hand side sum is $ka+(n-k)(a+1)$, that is, $na+n-k$.