Lucas/Fibonacci Proof By Induction?

Question

Hey guys,

I've been having so many troubles with this problem and I'm hoping that you could give me some guidance. The problem is to prove the following by induction:

F(2n) = F(n)*L(n)

Here, F(n) is the Fibonacci Sequence and L(n) is the Lucas Sequence. I have about 3 paths I've gone down and none of them end well. If it helps you, I have a side proof to show that L(n) = F(n-1) + F(n+1), so that can be used if needed.

I appreciate any help you can give with this problem. 

Thanks!

kb · Accepted Answer

'll assume that F(2n) = F^2(n+1) - F^2(n-1).

Then, this is easy:
F(2n) = F^2(n+1) - F^2(n-1)
.........= (F(n+1) - F(n-1)) * (F(n+1) + F(n-1))
.........= F(n) * (F(n+1) + F(n-1)), via Fibonacci recurrence
.........= F(n) * L(n), using your 'side relation'.
-------------------
Link to a proof of F(2n) = F^2(n+1) - F^2(n-1):
http://ph.answers.yahoo.com/question/index?qid=20120707152119AAlPKqP

I hope this helps!

? · Answer

3^m-1>5m for every positive integer m>or= to four i) begin with m = four 3^(m-1) = 3^3 = 27 > 5(four) = 20 ii) think genuine for m = okay show that it's actual for m = okay+1 expect three^(k - 1) > 5k let m = k+1 three^(m - 1) = three^okay = three(3^(k-1)) > 3(5k) = 15k = 5k + 10k > 5k + 5 ` ` ` ` considering that k > 1 = 5(ok+1)

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Lucas/Fibonacci Proof By Induction?

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