Question

$f(n)=[ \frac{n^2}{3} ]+[ \frac{(n+1)^2}{3} ] +[\frac{(n+2)^2}{3} ] \\ Aratati-mi~si~mie~ca~f(3k)=(3k+1)^2$

Answer 1

$f(3k)=\Big[\frac{(3k)^2}{3}\Big]+\Big[\frac{(3k+1)^2}{3}\Big]+\Big[\frac{(3k+2)^2}{3}\Big] \\ \\ =\Big[\frac{(3k)^2}{3}\Big]+\Big[\frac{(3k)^2+6k+1}{3}\Big]+\Big[\frac{(3k)^2+12k+4}{3}\Big] \\ \\ =\Big[\frac{9k^2}{3}\Big]+\Big[\frac{9k^2+6k+1}{3}\Big]+\Big[\frac{9k^2+12k+4}{3}\Big] \\ \\ =[3k^2]+\Big[3k^2+2k+\frac{1}{3}\Big]+\Big[3k^2+4k+\frac{4}{3}\Big] \\ \\ =3k^2+(3k^2+2k)+(3k^2+4k+1) \\ \\ =3k^2+3k^2+2k+3k^2+4k+1 \\ \\ =9k^2+6k+1 \\ \\ =(3k)^2+2\cdot3k\cdot1+1^2 \\ \\ =(3k+1)^2$