## mathmath333 one year ago greatest integer function

1. mathmath333

if $$\large \lfloor{\rfloor}\rightarrow$$ is Greatest integer function then find value of \large \color{black}{\begin{align} \lfloor{\sqrt1\rfloor}+\lfloor{\sqrt2\rfloor}+\lfloor{\sqrt3\rfloor}+\cdots+\lfloor{\sqrt{50}\rfloor}=\hspace{1.5em}\\~\\ \end{align}}

2. geerky42

Well, here's a little 'brute force' way to solve it: We can see that in between 1 and 4, there are 3 integers So $$\lfloor \sqrt{1}\rfloor+\lfloor \sqrt{2}\rfloor+\lfloor \sqrt{3}\rfloor = 3\cdot1 = 3$$ Between 4 and 9, there are 5 integers So $$\lfloor \sqrt{4}\rfloor+\lfloor \sqrt{5}\rfloor+\lfloor \sqrt{6}\rfloor+\lfloor \sqrt{7}\rfloor+\lfloor \sqrt{8}\rfloor = 5\cdot 2= 10$$ One more... Between 9 and 16, there are 7 integers So $$\lfloor \sqrt{9}\rfloor+\lfloor \sqrt{10}\rfloor+\lfloor \sqrt{11}\rfloor+\lfloor \sqrt{12}\rfloor+\lfloor \sqrt{13}\rfloor+\lfloor \sqrt{14}\rfloor+\lfloor \sqrt{15}\rfloor = 7\cdot 3= 21$$ See the pattern? Generalize it to n, we have that for $$n^2\le~k~<(n+1)^2$$, we have $\sum_{n^2\le~k~<(n+1)^2} \lfloor \sqrt{~k~}\rfloor = (2n+1)n$ So here, you have $\lfloor{\sqrt1\rfloor}+\lfloor{\sqrt2\rfloor}+\lfloor{\sqrt3\rfloor}+\cdots+\lfloor{\sqrt{50}\rfloor}\\~\\~~~~~= \sum_{1\le~k~<4} \lfloor \sqrt{~k~}\rfloor +\sum_{4\le~k~<9} \lfloor \sqrt{~k~}\rfloor +\sum_{9\le~k~<16} \lfloor \sqrt{~k~}\rfloor \\~\\~~~~~~~~~~~~~~~~~~~~~+\sum_{16\le~k~<25} \lfloor \sqrt{~k~}\rfloor +\sum_{25\le~k~<36} \lfloor \sqrt{~k~}\rfloor+\sum_{36\le~k~<49} \lfloor \sqrt{~k~}\rfloor +\lfloor\sqrt{49}\rfloor+\lfloor\sqrt{50}\rfloor\\~\\~\\ = (3)1+(5)2+(7)3+(9)4+(11)5+(13)6+\lfloor\sqrt{49}\rfloor+\lfloor\sqrt{50}\rfloor\\~\\=3+10+21+36+55+78 + 7+7\\~\\=\boxed{217}$