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# Telescoping series

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Compute $$\frac{\lfloor \sqrt{1} \rfloor \cdot \lfloor \sqrt{3} \rfloor \cdot \lfloor \sqrt{5} \rfloor \dotsm \lfloor \sqrt{2015} \rfloor}{\lfloor \sqrt{2} \rfloor \cdot \lfloor \sqrt{4} \rfloor \cdot \lfloor \sqrt{6} \rfloor \dotsm \lfloor \sqrt{2016} \rfloor}.$$

Dec 2, 2018
edited by Guest  Dec 2, 2018

#1
0

deleted. Sorry, misread the "floor function"

Dec 2, 2018
edited by Guest  Dec 3, 2018
#2
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deleted.

Dec 2, 2018
edited by Guest  Dec 2, 2018
edited by Guest  Dec 3, 2018
#3
+1

Note that

2^4 = 16

3^4 = 81

4^4 = 256

5^4 = 625

6^4 = 1296

7^4 = 2401

So each floor from (1)^.25 to ( 15)^.25  = 1

And floor (16)^.25 = 2

And each  floor  from (17)^.25 to (80)^.25 = 2

And floor (81)^.25 = 3

And each floor  from (82)^.25 to (255)^.25 = 3

And floor (256)^.25 = 4

And each floor  from (257)^.25 to (624)^.25 = 4

And floor (625)^.25 = 5

And each floor  from (626)^.25 to (1295)^.25 = 5

And floor (1296)^.25 = 6

And each floor  from (1297)^.25 to (2016)^.25 =  6

So...note the pattern.....

(1)^8 *  (2)^32 * (3)^88 * (4)^184 * ....               1 * 3 * ......

______________________________        =  ___________

(1)^7 * (2)^33 * (3)^87* (4)^185 * .......               2 * 4 * .....

1 *  3 *  5 *

__________  =

2 * 4 * 6

15

____ =

48

5

__

16   Dec 2, 2018
edited by CPhill  Dec 3, 2018
edited by CPhill  Dec 3, 2018
#4
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How can the top be greater than the bottom though

Guest Dec 3, 2018
#5
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Sorry.....I shouldn't have included "7' in the numerator...we only have a floor of 6 on (2016)^.25

I believe tthe final answer is correct....thanks for noticing my error...   CPhill  Dec 3, 2018
edited by CPhill  Dec 3, 2018
#6
+2

I think your calculation is incorrect, the answer I got is $$\frac{1}{\lfloor \sqrt{2016} \rfloor}$$

HelloWorld  Dec 3, 2018
edited by HelloWorld  Dec 3, 2018