Lucas Number Proof By Induction

**Kiwi_Dave** · July 26th 2008, 12:48 AM

I am having some trouble proving the relationship

$F_n$ $L_n$ = $F_{2n}$

where $F_n$ , $L_n$ are the Fibronachi and Lucas numbers respectively.

Oh, and how should I have written this in Laytex?

**Kiwi_Dave** · September 15th 2008, 07:07 PM

Can anyone give this a go? I am expecting to solve it by induction. But I can see no way of making the leap from numbers like F_n all the way up to F_2n.

The inductive hypothesis doesn't apply to F_(2n+1) so I can see no way to proceed.

**Moo** · September 16th 2008, 12:23 AM

Hello,

Try strong induction Mathematical induction - Wikipedia, the free encyclopedia

And it's Fibonacci, not Fibronachi

**NonCommAlg** · September 16th 2008, 04:08 PM

Originally Posted by Kiwi_Dave

I am having some trouble proving the relationship

$F_n$ $L_n$ = $F_{2n}$

where $F_n$ , $L_n$ are the Fibronachi and Lucas numbers respectively.

well, it's not a very straightforward induction! it's clearly true for n = 1, 2, 3. so assuming

that n > 3 and that the claim is true for all $1 \leq k \leq n,$ we'll have:

$F_{n+1}L_{n+1}=(F_n+F_{n-1})(L_n+L_{n-1})=F_nL_n+F_nL_{n-1}+F_{n-1}L_n+F_{n-1}L_{n-1}$

$=F_{2n}+F_{2n-2}+F_nL_{n-1}+F_{n-1}L_n=F_{2n}+F_{2n-2}+ F_n(L_n - L_{n-2}) + F_{n-1}(L_{n-1}+L_{n-2})$

$=2F_{2n}+2F_{2n-2}-L_{n-2}(F_n-F_{n-1})=2F_{2n}+2F_{2n-2}-L_{n-2}F_{n-2}=2F_{2n}+2F_{2n-2}-F_{2n-4}$

$=2F_{2n}+2F_{2n-2}-(F_{2n-2}-F_{2n-3})=2F_{2n}+F_{2n-2}+F_{2n-3}=2F_{2n}+F_{2n-1}=F_{2n}+F_{2n+1}$

$=F_{2n+2}. \ \clubsuit$

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