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PikachuMaxwell
05/04/20 12:17:01 AM
#1:


...13, 21, 34, 55, 89....

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andri_g
05/04/20 1:56:11 AM
#2:


... 144, 233, 377, 610, 1087...
... 1697, 2784, 4481, 7165, 11646...

<P

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PikachuMaxwell
05/04/20 2:14:55 AM
#3:


andri_g posted...
... 144, 233, 377, 610, 1087...
... 1697, 2784, 4481, 7165, 11646...

<P

Yay!

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TheDarkCircle
05/04/20 2:16:20 AM
#4:


i remember learning about this shit in math class in middle school, wtf was the point of this again?
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PikachuMaxwell
05/04/20 2:18:16 AM
#5:


TheDarkCircle posted...
i remember learning about this shit in math class in middle school, wtf was the point of this again?

I dunno...but it's fun to know these things (for me at least!)

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Pogo_Marimo
05/04/20 2:30:20 AM
#6:




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I presume my time here in my darkblack dragondark steel-obliterating solitude has come to its end as well.
http://www.last.fm/user/Pogo92
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ChocoboMog123
05/04/20 2:38:58 AM
#7:


a(n) = (((1+sqrt(5))/2)^n-((1-sqrt(5))/2)^n))/sqrt(5)

Proof by induction:

a(1) = (((1+sqrt(5))/2)^1-((1-sqrt(5))/2)^1))/sqrt(5) = 1 (plugged it into wolfram)
a(2) = ... = 1
Show that:
a(n+1) = (((1+sqrt(5))/2)^(n+1)-((1-sqrt(5))/2)^(n+1)))/sqrt(5)
a(n+1)=a(n)+a(n-1) by definition of Fibonacci sequence
a(n+1) = [(((1+sqrt(5))/2)^n-((1-sqrt(5))/2)^n))/sqrt(5)] + [(((1+sqrt(5))/2)^(n-1)-((1-sqrt(5))/2)^(n-1)))/sqrt(5)]
algebra is a bit messy, but it should work out.

QED

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PikachuMaxwell
05/04/20 2:40:03 AM
#8:


ChocoboMog123 posted...
a(n) = (((1+sqrt(5))/2)^n-((1-sqrt(5))/2)^n))/sqrt(5)

Proof by induction:

a(1) = (((1+sqrt(5))/2)^1-((1-sqrt(5))/2)^1))/sqrt(5) = 1 (plugged it into wolfram)
a(2) = ... = 1
Show that:
a(n+1) = (((1+sqrt(5))/2)^(n+1)-((1-sqrt(5))/2)^(n+1)))/sqrt(5)
a(n+1)=a(n)+a(n-1) by definition of Fibonacci sequence
a(n+1) = [(((1+sqrt(5))/2)^n-((1-sqrt(5))/2)^n))/sqrt(5)] + [(((1+sqrt(5))/2)^(n-1)-((1-sqrt(5))/2)^(n-1)))/sqrt(5)]
algebra is a bit messy, but it should work out.

QED

Ah, cool!

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