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Math Help - Fibonacci Sequence property

  1. #1
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    Fibonacci Sequence property

    Using normal properties of Fibonacci Numbers Prove that:
    (F(n+1)^2)+(F(n)^2)= F(2n+1).
    We see that F(3)=2 and F(4)=3 and so F(7)=13 so it is true
    Not sure where to start here and any help would be appreciated.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by MatsSundin View Post
    Using normal properties of Fibonacci Numbers Prove that:
    (F(n+1)^2)+(F(n)^2)= F(2n+1).
    We see that F(3)=2 and F(4)=3 and so F(7)=13 so it is true
    Not sure where to start here and any help would be appreciated.
    i suppose you are doing induction.

    so, you must now assume P(n) is true and try to show that P(n + 1) must be true.

    of course, here, P(n) is " F_{n + 1}^2 + F_n^2 = F_{2n + 1} for all integers n \ge 3"
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  3. #3
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    I proved this identity back in school using Binet's formula(without induction) and some identities.

    Do you have to use induction?. That is a good way to go, but I like this method I stumbled onto a while back while studying Fibonacci's for a class called Seminars in Mathematics. I doubt if it's original, but it works.

    I will use x=\frac{1+\sqrt{5}}{2}, \;\ y=\frac{1-\sqrt{5}}{2}

    Take note that xy=-1 and x^{2}-\sqrt{5}x+1=0 and y^{2}+\sqrt{5}y+1=0

    We will use these later.

    From Binet's formula: \frac{x^{n}-y^{n}}{\sqrt{5}}

    We can sub it in our formula to prove:

    \frac{1}{5}\left[(x^{n}-y^{n})^{2}+(x^{n+1}-y^{n+1})^{2}-\sqrt{5}(x^{2n+1}-y^{2n+1})\right]

    Expand:

    \frac{1}{5}\left[x^{2n}-2x^{n}y^{n}+y^{2n}+x^{2n+2}-2x^{n+1}y^{n+1}+y^{2n+2}-\sqrt{5}x^{2n+1}+\sqrt{5}y^{2n+1}\right]

    Group and factor:

    \frac{1}{5}\left[x^{2n}(\underbrace{x^{2}-\sqrt{5}x+1}_{\text{equals 0}})+y^{2n}(\underbrace{y^{2}+\sqrt{5}y+1}_{\text{  equals 0}})-2(\underbrace{xy}_{\text{-1}})^{n}(1+xy)\right]

    Take note that everything inside the brackets is 0, due to the identities previously mentioned.

    We get:

    \frac{1}{5}\left[x^{2n}(0)+y^{2n}(0)-2(-1)(0)\right]=0

    \therefore, f^{2}_{n}+f^{2}_{n+1}=f_{2n+1}

    You can show various identities using these tools. They can be handy to know.
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  4. #4
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    No induction is not needed, it can be proven any way possible. Your solution is quite intuitive and impressive thank you!
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