mathmath333
  • mathmath333
greatest integer function
Mathematics
schrodinger
  • schrodinger
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

mathmath333
  • mathmath333
if \(\large \lfloor{\rfloor}\rightarrow\) is Greatest integer function then find value of \(\large \color{black}{\begin{align} \lfloor{\sqrt1\rfloor}+\lfloor{\sqrt2\rfloor}+\lfloor{\sqrt3\rfloor}+\cdots+\lfloor{\sqrt{50}\rfloor}=\hspace{1.5em}\\~\\ \end{align}}\)
geerky42
  • 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 \[\lfloor{\sqrt1\rfloor}+\lfloor{\sqrt2\rfloor}+\lfloor{\sqrt3\rfloor}+\cdots+\lfloor{\sqrt{50}\rfloor}\\~\\~~~~~= \sum_{1\le~k~<4} \lfloor \sqrt{~k~}\rfloor +\sum_{4\le~k~<9} \lfloor \sqrt{~k~}\rfloor +\sum_{9\le~k~<16} \lfloor \sqrt{~k~}\rfloor \\~\\~~~~~~~~~~~~~~~~~~~~~+\sum_{16\le~k~<25} \lfloor \sqrt{~k~}\rfloor +\sum_{25\le~k~<36} \lfloor \sqrt{~k~}\rfloor+\sum_{36\le~k~<49} \lfloor \sqrt{~k~}\rfloor +\lfloor\sqrt{49}\rfloor+\lfloor\sqrt{50}\rfloor\\~\\~\\ = (3)1+(5)2+(7)3+(9)4+(11)5+(13)6+\lfloor\sqrt{49}\rfloor+\lfloor\sqrt{50}\rfloor\\~\\=3+10+21+36+55+78 + 7+7\\~\\=\boxed{217}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.