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 Post subject: How to prove F2n = Fn * Ln using induction
PostPosted: Mon, 28 Nov 2011 01:19:59 UTC 
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I need to show: F2n = Fn * Ln using induction

Where Fn stands for the nth Fibonacci number and Ln stands for the nth Lucas number.

This is problem #15 in chapter 1 of Number Theory by George Andrews. (I'm teaching myself for fun, this isn't homework.)

The book expects you to use induction. The only useful equations this chapter of the book (this is chapter 1) seems to provide are:

1. Fn = Fn-1 + Fn-2
2. F^2(n+1) - F(n)F(n+2) = (-1)^n
3. Ln = F(n+1) + F(n-1)
4. Ln = L(n-1 + L(n-2)

These were either given definitions, or previous problems that had to be solved, and can now be taken as givens.

I know that the quickest way to solve this is to substitute and then use the definition of F(m+n), but that definition is not a given here, so the author wants me to use one of the 4 givens.

I think I need another way to define F2n in terms of Fn, not just as F(2n-2) + F(2n-1).

Edit (Shadow): see post below


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 Post subject: Re: How to prove F2n = Fn * Ln using induction
PostPosted: Mon, 28 Nov 2011 01:31:18 UTC 
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Teknontheou wrote:
Where Fn stands for the nth Fibonacci number and Ln stands for the nth Lucas number.

This is problem #15 in chapter 1 of Number Theory by George Andrews. (I'm teaching myself for fun, this isn't homework.)

The book expects you to use induction. The only useful equations this chapter of the book (this is chapter 1) seems to provide are:

1. Fn = Fn-1 + Fn-2
2. F^2(n+1) - F(n)F(n+2) = (-1)^n
3. Ln = F(n+1) + F(n-1)
4. Ln = L(n-1 + L(n-2)

These were either given definitions, or previous problems that had to be solved, and can now be taken as givens.

I know that the quickest way to solve this is to substitute and then use the definition of F(m+n), but that definition is not a given here, so the author wants me to use one of the 4 givens.

I think I need another way to define F2n in terms of Fn, not just as F(2n-2) + F(2n-1).


Please post your question in the main body of your topic. I've edited your original post.

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 Post subject: Re: How to prove F2n = Fn * Ln using induction
PostPosted: Mon, 28 Nov 2011 01:42:39 UTC 
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I'm sure if you've seen a definition before it's fine to use it. Prove by induction doesn't necessarily mean EVERYTHING is induction.

In any case, F_{2n}=F_{n+1}^2-F_{n-1}^2=F_n(F_{n+1}+F_{n-1})=F_nL_n, so try proving the first equality by induction instead.

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 Post subject: Re: How to prove F2n = Fn * Ln using induction
PostPosted: Mon, 28 Nov 2011 01:56:17 UTC 
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Shadow wrote:
I'm sure if you've seen a definition before it's fine to use it. Prove by induction doesn't necessarily mean EVERYTHING is induction.

In any case, F_{2n}=F_{n+1}^2-F_{n-1}^2=F_n(F_{n+1}+F_{n-1})=F_nL_n, so try proving the first equality by induction instead.


Thanks. I'm going to play around with this some more.


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 Post subject: Re: How to prove F2n = Fn * Ln using induction
PostPosted: Mon, 28 Nov 2011 05:31:28 UTC 
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Teknontheou wrote:
Shadow wrote:
I'm sure if you've seen a definition before it's fine to use it. Prove by induction doesn't necessarily mean EVERYTHING is induction.

In any case, F_{2n}=F_{n+1}^2-F_{n-1}^2=F_n(F_{n+1}+F_{n-1})=F_nL_n, so try proving the first equality by induction instead.


Thanks. I'm going to play around with this some more.


Alternatively, prove that 2F_nL_n=F_{n+1}L_{n-1}+F_{n-1}L_{n+1}, which implies the result when you realise F_{N+2}=3F_N-F_{N-2} (apply this to the case N=2n).

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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