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# help

-3
53
4
+389

Evaluate$$\lfloor\sqrt{1}\rfloor + \lfloor\sqrt{2}\rfloor + \lfloor\sqrt{3}\rfloor + .... + \lfloor\sqrt{19}\rfloor$$

Mar 31, 2020

#1
+1948
+2

Hint:

This is rounding DOWN to the nearest whole number. For example, sqrt(2) would be 1

Mar 31, 2020
#2
+416
0

That is true.. Cal are you really in elementery school?

Mar 31, 2020
#4
+1948
+1

Yes. Why?

CalTheGreat  Mar 31, 2020
#3
+626
+2

Hellow Qwertyzz!

Do you know what floor function is? Just a reminder: Floor(3.9) = 3

Sqrt(1) = 1

Sqrt(2) = 1.414...

Sqrt(3) = 1.732...

Sqrt(4) = 2

Sqrt(5) = 2.236...

...

Sqrt (9) = 3

....

Sqrt(16) = 4

See the pattern?

The terms evaluate into an integer and only increases at the next perfect square.

Solving:

1 + 1 + 1 + 2 + 2 + 2 + 2 + 2 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 4 + 4 + 4 + 4

= Gonna let you do some work

Mar 31, 2020
edited by AnExtremelyLongName  Mar 31, 2020