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Evaluate $$\lfloor\sqrt{1}\rfloor + \lfloor\sqrt{2}\rfloor + \lfloor\sqrt{3}\rfloor + \dots + \lfloor\sqrt{29}\rfloor$$

 Jun 7, 2022

Best Answer 

 #2
avatar+2448 
0

\(\lfloor \sqrt 1 \rfloor, \lfloor \sqrt2 \rfloor, \lfloor \sqrt 3 \rfloor\) These 3 terms each equal 1

 

\(\lfloor \sqrt4 \rfloor, \lfloor \sqrt5 \rfloor \cdot \cdot \cdot ,\lfloor \sqrt8 \rfloor\) These 5 terms each equal 2

 

\(\lfloor \sqrt{9} \rfloor, \lfloor \sqrt{10} \rfloor \cdot \cdot \cdot ,\lfloor \sqrt{15} \rfloor\) These 7 terms each equal 3

 

\(\lfloor \sqrt{16} \rfloor, \lfloor \sqrt{17} \rfloor \cdot \cdot \cdot ,\lfloor \sqrt{24} \rfloor\) These 9 terms each equal 4

 

\(\lfloor \sqrt{25} \rfloor, \lfloor \sqrt{26} \rfloor \cdot \cdot \cdot ,\lfloor \sqrt{29} \rfloor\) These 5 terms each equal 5

 

Thus, the sum is \((5 \times 5) + (9 \times 4) + (7 \times 3) + (5 \times 2) + (3 \times 1) = \color{brown}\boxed{95}\)

 Jun 7, 2022
 #1
avatar+367 
+2

Hey there, Guest!

 

So I just wrote out all the square roots:

 

\(\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{5}+\sqrt{6}+\sqrt{7}+\sqrt{8}+\sqrt{9}+\sqrt{10}\sqrt{11}+\sqrt{12}+\sqrt{13}+\sqrt{14}+\sqrt{15}+\sqrt{16}+\sqrt{17}+\sqrt{18}+\sqrt{19}+\sqrt{20}+\sqrt{21}+\sqrt{22}+\sqrt{23}+\sqrt{24}+\sqrt{25}+\sqrt{26}+\sqrt{27}+\sqrt{28}+\sqrt{29}\)

 

It's probably not the smartest idea, but it works.

Then I simplified all the possible squares.

 

\(15+√110+√13+√14+√15+√17+√19+6√2+√21+√22+√23+√26+√29+6√3+3√5+3√6+3√7\)

 

Then you can simplify that into a decimal and you get 110.61.

 

And yes, I used a calculator for the last part, because who wants to do that by hand?

Not me, haha.

 

Hope this helped! :)

( ゚д゚)つ Bye

 Jun 7, 2022
 #2
avatar+2448 
0
Best Answer

\(\lfloor \sqrt 1 \rfloor, \lfloor \sqrt2 \rfloor, \lfloor \sqrt 3 \rfloor\) These 3 terms each equal 1

 

\(\lfloor \sqrt4 \rfloor, \lfloor \sqrt5 \rfloor \cdot \cdot \cdot ,\lfloor \sqrt8 \rfloor\) These 5 terms each equal 2

 

\(\lfloor \sqrt{9} \rfloor, \lfloor \sqrt{10} \rfloor \cdot \cdot \cdot ,\lfloor \sqrt{15} \rfloor\) These 7 terms each equal 3

 

\(\lfloor \sqrt{16} \rfloor, \lfloor \sqrt{17} \rfloor \cdot \cdot \cdot ,\lfloor \sqrt{24} \rfloor\) These 9 terms each equal 4

 

\(\lfloor \sqrt{25} \rfloor, \lfloor \sqrt{26} \rfloor \cdot \cdot \cdot ,\lfloor \sqrt{29} \rfloor\) These 5 terms each equal 5

 

Thus, the sum is \((5 \times 5) + (9 \times 4) + (7 \times 3) + (5 \times 2) + (3 \times 1) = \color{brown}\boxed{95}\)

BuilderBoi Jun 7, 2022

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