Denoting $[x]$ the integer part of the real number $x$, prove:
$$\left [ \frac{k(k+1)}{4k-2} \right ]=\left [ \frac{k+1}{4}\right ],\;\;k\in\mathbb{Z}.$$
Comment : I found the problem while leafing through the magazine INTEGERS (Electronic Journal of Combinatorial Number Theory). In volume 20, year 2020, an article by PALKA Ryszard and TRUKSZYN Zdzislsv with the title FINITE SUMS OF THE FLOOR FUNCTIONS SERIES appears. (See author index.) On page #A41:1 the formula (with number (5) in the text) that is the subject of our post is mentioned. (The pages are individually numbered for each article, starting with 1.) Among the chief editors is Florian LUCA.
