+0  
 
-3
53
4
avatar+389 

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

 Mar 31, 2020
 #1
avatar+1948 
+2

Hint:

 

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

 Mar 31, 2020
 #2
avatar+416 
0

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

 Mar 31, 2020
 #4
avatar+1948 
+1

Yes. Why?

CalTheGreat  Mar 31, 2020
 #3
avatar+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

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