Fibonacci and lucas series.

  • Thread starter Suk-Sci
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  • #1
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Main Question or Discussion Point

Fibonacci and lucas series.......

Let a1,a2,a3......,an be the numbers of fibonacci series.....
Let b1,b2.........bn be the number of lucas series.

bn=an-1 + an+1 for n[tex]\geq[/tex]2

T.P.T : a2n=an*bn
 

Answers and Replies

  • #2
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T.P.T : a2n=an*bn
Keep in mind the following two identities...

(Lucas_(n-1) + Lucas_(n+1))/5 = Fibonacci_n
Lucas_n = (Golden Ratio)^n + (-1)^n(Golden Ratio)^-n

... where the Golden Ratio = ((sqrt 5) + 1)/2
 
  • #3
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Hi, Suk-Sci,
you can substitute the given expression for the b's into the equation you want to prove; then you will have something only in terms of a's, that you can prove using induction.

If you want more help, try to show what you have done so far; that helps us help you. :)
 

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