Prove that if a and b are relatively prime, then
$$\sum_{n=1}^{a-1} \left\lfloor \frac{nb}{a}\right\rfloor = \frac{(a - 1)(b - 1)}{2}$$
My attempt was:
We have: $$\sum_{i=1}^{n-1} i = \frac{n(n - 1)}{2}$$
Then, $$\sum_{n=1}^{a-1} \left\lfloor \frac{nb}{a}\right\rfloor = \left\lfloor \frac{a(a - 1)b}{2a}\right\rfloor$$
Could I apply the summation formula for floor function like above? Am I in the right track?
Thanks,
Chan