MathDB
Hungary-Israel Binational 2008\4

Source: floor function identity

November 5, 2008
floor functionalgebra proposedalgebra

Problem Statement

Prove that: \sum_{i\equal{}1}^{n^2} \lfloor \frac{i}{3} \rfloor\equal{} \frac{n^2(n^2\minus{}1)}{6} For all n∈N n \in N.