Prove Fibonacci Sequence Property: $x^2_n + x^2_{n+1}=x_{2n+1}$












2












$begingroup$


For Fibonacci Sequence, We know that the recursive difference equation is:



$$
x_{n+2} = x_n + x_{n+1} n geq 0
$$



And that the closed form solution is:



$$
x_n = frac{1}{sqrt{5}}left[ left(frac{1+sqrt{5}}{2}right)^{n} - left(frac{1-sqrt{5}}{2}right)^{n} right] n geq 0
$$



But how to prove that this fibonacci property is true using this information?:



$$x^2_n + x^2_{n+1}=x_{2n+1}$$



Hint: substitute the closed form expression for the fibonacci sequence into difference equation and verify that its true.



well...I've tried this at least 3 different ways of doing exactly as hinted without much luck...the equation always blows up on me. Any ideas how to prove it?






discrete-mathematics fibonacci-numbers







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  • 1




    $begingroup$
    Can you show one or more of those 3 different ways? It is hard to guess at what you're doing wrong when we don't know what you're doing.
    $endgroup$
    – Henning Makholm
    Jan 7 at 23:43










  • $begingroup$
    Hint: with $x_0=0,x_1=1$, prove by induction over $m$ that $x_p=x_{m+1}x_{p-m}+x_mx_{p-m-1}$.
    $endgroup$
    – Mindlack
    Jan 8 at 0:06
















2












$begingroup$


For Fibonacci Sequence, We know that the recursive difference equation is:



$$
x_{n+2} = x_n + x_{n+1} n geq 0
$$



And that the closed form solution is:



$$
x_n = frac{1}{sqrt{5}}left[ left(frac{1+sqrt{5}}{2}right)^{n} - left(frac{1-sqrt{5}}{2}right)^{n} right] n geq 0
$$



But how to prove that this fibonacci property is true using this information?:



$$x^2_n + x^2_{n+1}=x_{2n+1}$$



Hint: substitute the closed form expression for the fibonacci sequence into difference equation and verify that its true.



well...I've tried this at least 3 different ways of doing exactly as hinted without much luck...the equation always blows up on me. Any ideas how to prove it?






discrete-mathematics fibonacci-numbers







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Can you show one or more of those 3 different ways? It is hard to guess at what you're doing wrong when we don't know what you're doing.
    $endgroup$
    – Henning Makholm
    Jan 7 at 23:43










  • $begingroup$
    Hint: with $x_0=0,x_1=1$, prove by induction over $m$ that $x_p=x_{m+1}x_{p-m}+x_mx_{p-m-1}$.
    $endgroup$
    – Mindlack
    Jan 8 at 0:06














2












2








2


1



$begingroup$


For Fibonacci Sequence, We know that the recursive difference equation is:



$$
x_{n+2} = x_n + x_{n+1} n geq 0
$$



And that the closed form solution is:



$$
x_n = frac{1}{sqrt{5}}left[ left(frac{1+sqrt{5}}{2}right)^{n} - left(frac{1-sqrt{5}}{2}right)^{n} right] n geq 0
$$



But how to prove that this fibonacci property is true using this information?:



$$x^2_n + x^2_{n+1}=x_{2n+1}$$



Hint: substitute the closed form expression for the fibonacci sequence into difference equation and verify that its true.



well...I've tried this at least 3 different ways of doing exactly as hinted without much luck...the equation always blows up on me. Any ideas how to prove it?






discrete-mathematics fibonacci-numbers







share|cite|improve this question









$endgroup$




For Fibonacci Sequence, We know that the recursive difference equation is:



$$
x_{n+2} = x_n + x_{n+1} n geq 0
$$



And that the closed form solution is:



$$
x_n = frac{1}{sqrt{5}}left[ left(frac{1+sqrt{5}}{2}right)^{n} - left(frac{1-sqrt{5}}{2}right)^{n} right] n geq 0
$$



But how to prove that this fibonacci property is true using this information?:



$$x^2_n + x^2_{n+1}=x_{2n+1}$$



Hint: substitute the closed form expression for the fibonacci sequence into difference equation and verify that its true.



well...I've tried this at least 3 different ways of doing exactly as hinted without much luck...the equation always blows up on me. Any ideas how to prove it?






discrete-mathematics fibonacci-numbers




discrete-mathematics fibonacci-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Jan 7 at 23:36









DiscreteMathDiscreteMath

132




132








  • 1




    $begingroup$
    Can you show one or more of those 3 different ways? It is hard to guess at what you're doing wrong when we don't know what you're doing.
    $endgroup$
    – Henning Makholm
    Jan 7 at 23:43










  • $begingroup$
    Hint: with $x_0=0,x_1=1$, prove by induction over $m$ that $x_p=x_{m+1}x_{p-m}+x_mx_{p-m-1}$.
    $endgroup$
    – Mindlack
    Jan 8 at 0:06














  • 1




    $begingroup$
    Can you show one or more of those 3 different ways? It is hard to guess at what you're doing wrong when we don't know what you're doing.
    $endgroup$
    – Henning Makholm
    Jan 7 at 23:43










  • $begingroup$
    Hint: with $x_0=0,x_1=1$, prove by induction over $m$ that $x_p=x_{m+1}x_{p-m}+x_mx_{p-m-1}$.
    $endgroup$
    – Mindlack
    Jan 8 at 0:06








1




1




$begingroup$
Can you show one or more of those 3 different ways? It is hard to guess at what you're doing wrong when we don't know what you're doing.
$endgroup$
– Henning Makholm
Jan 7 at 23:43




$begingroup$
Can you show one or more of those 3 different ways? It is hard to guess at what you're doing wrong when we don't know what you're doing.
$endgroup$
– Henning Makholm
Jan 7 at 23:43












$begingroup$
Hint: with $x_0=0,x_1=1$, prove by induction over $m$ that $x_p=x_{m+1}x_{p-m}+x_mx_{p-m-1}$.
$endgroup$
– Mindlack
Jan 8 at 0:06




$begingroup$
Hint: with $x_0=0,x_1=1$, prove by induction over $m$ that $x_p=x_{m+1}x_{p-m}+x_mx_{p-m-1}$.
$endgroup$
– Mindlack
Jan 8 at 0:06










4 Answers
4






active

oldest

votes


















7












$begingroup$

The 'standard proof' uses strong induction, and I'll leave you to figure that out.



Let $varphi$ and $psi$ represent the two roots of $x^2-x-1=0$ where $varphi > psi$ (i.e. $varphi,psi=frac{1pm sqrt{5}}{2}$). The given closed form solution is $$F_n=frac{varphi ^n-psi ^n}{sqrt{5}}$$ so we know $$F_n^2=frac{varphi ^{2n}-2(varphi psi)^n+psi ^{2n} }{5}$$ and $$F_{n+1}^2=frac{varphi ^{2n+2}-2(varphi psi )^{n+1}+psi ^{2n+2}}{5}$$ so $$F_n^2+F_{n+1}^2=frac{varphi ^{2n}-2(varphi psi)^n+psi ^{2n}+varphi ^{2n+2}-2(varphi psi )^{n+1}+psi ^{2n+2} }{5}$$But we know that $varphi psi =-1$, so either $(varphipsi)^n=1$ and $(varphipsi)^{n+1}=-1$ or $(varphipsi)^n=-1$ and $varphipsi)^{n+1}=1$. Either way, the two terms cancel each other, so$$F_n^2+F_{n+1}^2=frac{varphi^{2n}(1+varphi ^2)+psi ^{2n}(1+psi^2)}{5}$$It can be easily seen that $frac{1+varphi^2}{sqrt{5}}=varphi$ and $frac{1+psi^2}{sqrt{5}}=-psi$ so the above becomes begin{align*}F_n^2+F_{n+1}^2&=frac{varphi^{2n}(1+varphi ^2)+psi ^{2n}(1+psi^2)}{5}\ &=frac{varphi^{2n}cdot varphisqrt{5} + psi^{2n}cdot (-psisqrt{5})}{5} \ &=frac{sqrt{5}(varphi^{2n+1}-psi^{2n+1})}{5} \ &=frac{varphi^{2n+1}-psi^{2n+1}}{sqrt{5}} \ &= F_{2n+1}end{align*}completing the proof.






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$endgroup$









  • 1




    $begingroup$
    This seems to be the approach being hinted at. From OP's perspective, I think the insight that was missing was that we should expand things in terms of $varphi$ and $psi$, rather than in the way one typically "expands and simplifies".
    $endgroup$
    – Omnomnomnom
    Jan 8 at 0:10





















7












$begingroup$

For what it's worth, there's a nice approach using matrices. We can characterize the recurrence by
$$
pmatrix{x_{n+1}\x_n} = pmatrix{1&1\1&0}pmatrix{x_{n}\x_{n-1}}, quad n geq 1
$$
Let $F = pmatrix{1&1\1&0}$. As a consequence of this characterization, we end up with the formula
$$
F^n = pmatrix{x_{n+1} & x_n\ x_n & x_{n-1}}
$$
We now compute
$$
(F^n)^2 = pmatrix{x_{n+1} & x_n\ x_n & x_{n-1}}^2 = pmatrix{x_{n+1}^2 + x_n^2 & x_{n+1}x_n + x_n x_{n-1}\
x_{n+1}x_n + x_n x_{n-1} & x_n^2 + x_{n-1}^2}
$$
On the other hand,
$$
(F^n)^2 = F^{2n} = pmatrix{x_{2n+1} & x_{2n}\ x_{2n} & x_{2n-1}}
$$
Because the two matrices are equal, their upper-left entries are equal, which is exactly the desired conclusion.






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    0












    $begingroup$

    $x_{2n+1}=x_{2n}+x_{2n-1}=x_2x_{2n}+x_1x_{2n-1}=x_2(x_{2n-1}+x_{2n-2})+x_1x_{2n-1}=(x_1+x_2)x_{2n-1}+x_2x_{2n-2}=x_3x_{2n-1}+x_2x_{2n-2}=x_4x_{2n-2}+x_3x_{2n-3}=cdots =x_xx_n+x_{n+1}x_{n+1}=x_n^2+x_{n+1}^2.$






    share|cite|improve this answer











    $endgroup$





















      -3












      $begingroup$

      Based on the title of the topic and the hint, you need to plug the Closed form expression into your equality



      $$x_{n}^2+x_{n+1}^2-x_{2n+1}$$



      and make sure it equals zero identically.



      Best way is to taky any CAS and plug the figures. For Wolfram Mathematica




      x[n_] := 1/Sqrt[5]*(((1 + Sqrt[5])/2)^n - ((1 - Sqrt[5])/2)^n);



      x[a]^2 + x[a + 1]^2 - x[2*a + 1]



      Simplify[Out]




      Then you get something like this



      $$frac{1}{5} left(left(frac{2}{sqrt{5}+1}right)^{-a-1}-left(frac{1}{2} left(1-sqrt{5}right)right)^{a+1}right)^2+frac{1}{5} left(left(frac{1}{2} left(sqrt{5}+1right)right)^a-left(frac{1}{2} left(1-sqrt{5}right)right)^aright)^2-frac{left(frac{2}{sqrt{5}+1}right)^{-2 a-1}-left(frac{1}{2} left(1-sqrt{5}right)right)^{2 a+1}}{sqrt{5}}$$



      You need to carefully open up all brackets, cancel out the terms to get zero.






      share|cite|improve this answer









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      • 2




        $begingroup$
        relevant comic
        $endgroup$
        – Omnomnomnom
        Jan 8 at 1:44











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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7












      $begingroup$

      The 'standard proof' uses strong induction, and I'll leave you to figure that out.



      Let $varphi$ and $psi$ represent the two roots of $x^2-x-1=0$ where $varphi > psi$ (i.e. $varphi,psi=frac{1pm sqrt{5}}{2}$). The given closed form solution is $$F_n=frac{varphi ^n-psi ^n}{sqrt{5}}$$ so we know $$F_n^2=frac{varphi ^{2n}-2(varphi psi)^n+psi ^{2n} }{5}$$ and $$F_{n+1}^2=frac{varphi ^{2n+2}-2(varphi psi )^{n+1}+psi ^{2n+2}}{5}$$ so $$F_n^2+F_{n+1}^2=frac{varphi ^{2n}-2(varphi psi)^n+psi ^{2n}+varphi ^{2n+2}-2(varphi psi )^{n+1}+psi ^{2n+2} }{5}$$But we know that $varphi psi =-1$, so either $(varphipsi)^n=1$ and $(varphipsi)^{n+1}=-1$ or $(varphipsi)^n=-1$ and $varphipsi)^{n+1}=1$. Either way, the two terms cancel each other, so$$F_n^2+F_{n+1}^2=frac{varphi^{2n}(1+varphi ^2)+psi ^{2n}(1+psi^2)}{5}$$It can be easily seen that $frac{1+varphi^2}{sqrt{5}}=varphi$ and $frac{1+psi^2}{sqrt{5}}=-psi$ so the above becomes begin{align*}F_n^2+F_{n+1}^2&=frac{varphi^{2n}(1+varphi ^2)+psi ^{2n}(1+psi^2)}{5}\ &=frac{varphi^{2n}cdot varphisqrt{5} + psi^{2n}cdot (-psisqrt{5})}{5} \ &=frac{sqrt{5}(varphi^{2n+1}-psi^{2n+1})}{5} \ &=frac{varphi^{2n+1}-psi^{2n+1}}{sqrt{5}} \ &= F_{2n+1}end{align*}completing the proof.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        This seems to be the approach being hinted at. From OP's perspective, I think the insight that was missing was that we should expand things in terms of $varphi$ and $psi$, rather than in the way one typically "expands and simplifies".
        $endgroup$
        – Omnomnomnom
        Jan 8 at 0:10


















      7












      $begingroup$

      The 'standard proof' uses strong induction, and I'll leave you to figure that out.



      Let $varphi$ and $psi$ represent the two roots of $x^2-x-1=0$ where $varphi > psi$ (i.e. $varphi,psi=frac{1pm sqrt{5}}{2}$). The given closed form solution is $$F_n=frac{varphi ^n-psi ^n}{sqrt{5}}$$ so we know $$F_n^2=frac{varphi ^{2n}-2(varphi psi)^n+psi ^{2n} }{5}$$ and $$F_{n+1}^2=frac{varphi ^{2n+2}-2(varphi psi )^{n+1}+psi ^{2n+2}}{5}$$ so $$F_n^2+F_{n+1}^2=frac{varphi ^{2n}-2(varphi psi)^n+psi ^{2n}+varphi ^{2n+2}-2(varphi psi )^{n+1}+psi ^{2n+2} }{5}$$But we know that $varphi psi =-1$, so either $(varphipsi)^n=1$ and $(varphipsi)^{n+1}=-1$ or $(varphipsi)^n=-1$ and $varphipsi)^{n+1}=1$. Either way, the two terms cancel each other, so$$F_n^2+F_{n+1}^2=frac{varphi^{2n}(1+varphi ^2)+psi ^{2n}(1+psi^2)}{5}$$It can be easily seen that $frac{1+varphi^2}{sqrt{5}}=varphi$ and $frac{1+psi^2}{sqrt{5}}=-psi$ so the above becomes begin{align*}F_n^2+F_{n+1}^2&=frac{varphi^{2n}(1+varphi ^2)+psi ^{2n}(1+psi^2)}{5}\ &=frac{varphi^{2n}cdot varphisqrt{5} + psi^{2n}cdot (-psisqrt{5})}{5} \ &=frac{sqrt{5}(varphi^{2n+1}-psi^{2n+1})}{5} \ &=frac{varphi^{2n+1}-psi^{2n+1}}{sqrt{5}} \ &= F_{2n+1}end{align*}completing the proof.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        This seems to be the approach being hinted at. From OP's perspective, I think the insight that was missing was that we should expand things in terms of $varphi$ and $psi$, rather than in the way one typically "expands and simplifies".
        $endgroup$
        – Omnomnomnom
        Jan 8 at 0:10
















      7












      7








      7





      $begingroup$

      The 'standard proof' uses strong induction, and I'll leave you to figure that out.



      Let $varphi$ and $psi$ represent the two roots of $x^2-x-1=0$ where $varphi > psi$ (i.e. $varphi,psi=frac{1pm sqrt{5}}{2}$). The given closed form solution is $$F_n=frac{varphi ^n-psi ^n}{sqrt{5}}$$ so we know $$F_n^2=frac{varphi ^{2n}-2(varphi psi)^n+psi ^{2n} }{5}$$ and $$F_{n+1}^2=frac{varphi ^{2n+2}-2(varphi psi )^{n+1}+psi ^{2n+2}}{5}$$ so $$F_n^2+F_{n+1}^2=frac{varphi ^{2n}-2(varphi psi)^n+psi ^{2n}+varphi ^{2n+2}-2(varphi psi )^{n+1}+psi ^{2n+2} }{5}$$But we know that $varphi psi =-1$, so either $(varphipsi)^n=1$ and $(varphipsi)^{n+1}=-1$ or $(varphipsi)^n=-1$ and $varphipsi)^{n+1}=1$. Either way, the two terms cancel each other, so$$F_n^2+F_{n+1}^2=frac{varphi^{2n}(1+varphi ^2)+psi ^{2n}(1+psi^2)}{5}$$It can be easily seen that $frac{1+varphi^2}{sqrt{5}}=varphi$ and $frac{1+psi^2}{sqrt{5}}=-psi$ so the above becomes begin{align*}F_n^2+F_{n+1}^2&=frac{varphi^{2n}(1+varphi ^2)+psi ^{2n}(1+psi^2)}{5}\ &=frac{varphi^{2n}cdot varphisqrt{5} + psi^{2n}cdot (-psisqrt{5})}{5} \ &=frac{sqrt{5}(varphi^{2n+1}-psi^{2n+1})}{5} \ &=frac{varphi^{2n+1}-psi^{2n+1}}{sqrt{5}} \ &= F_{2n+1}end{align*}completing the proof.






      share|cite|improve this answer









      $endgroup$



      The 'standard proof' uses strong induction, and I'll leave you to figure that out.



      Let $varphi$ and $psi$ represent the two roots of $x^2-x-1=0$ where $varphi > psi$ (i.e. $varphi,psi=frac{1pm sqrt{5}}{2}$). The given closed form solution is $$F_n=frac{varphi ^n-psi ^n}{sqrt{5}}$$ so we know $$F_n^2=frac{varphi ^{2n}-2(varphi psi)^n+psi ^{2n} }{5}$$ and $$F_{n+1}^2=frac{varphi ^{2n+2}-2(varphi psi )^{n+1}+psi ^{2n+2}}{5}$$ so $$F_n^2+F_{n+1}^2=frac{varphi ^{2n}-2(varphi psi)^n+psi ^{2n}+varphi ^{2n+2}-2(varphi psi )^{n+1}+psi ^{2n+2} }{5}$$But we know that $varphi psi =-1$, so either $(varphipsi)^n=1$ and $(varphipsi)^{n+1}=-1$ or $(varphipsi)^n=-1$ and $varphipsi)^{n+1}=1$. Either way, the two terms cancel each other, so$$F_n^2+F_{n+1}^2=frac{varphi^{2n}(1+varphi ^2)+psi ^{2n}(1+psi^2)}{5}$$It can be easily seen that $frac{1+varphi^2}{sqrt{5}}=varphi$ and $frac{1+psi^2}{sqrt{5}}=-psi$ so the above becomes begin{align*}F_n^2+F_{n+1}^2&=frac{varphi^{2n}(1+varphi ^2)+psi ^{2n}(1+psi^2)}{5}\ &=frac{varphi^{2n}cdot varphisqrt{5} + psi^{2n}cdot (-psisqrt{5})}{5} \ &=frac{sqrt{5}(varphi^{2n+1}-psi^{2n+1})}{5} \ &=frac{varphi^{2n+1}-psi^{2n+1}}{sqrt{5}} \ &= F_{2n+1}end{align*}completing the proof.







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      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 8 at 0:04









      user574848user574848

      38817




      38817








      • 1




        $begingroup$
        This seems to be the approach being hinted at. From OP's perspective, I think the insight that was missing was that we should expand things in terms of $varphi$ and $psi$, rather than in the way one typically "expands and simplifies".
        $endgroup$
        – Omnomnomnom
        Jan 8 at 0:10
















      • 1




        $begingroup$
        This seems to be the approach being hinted at. From OP's perspective, I think the insight that was missing was that we should expand things in terms of $varphi$ and $psi$, rather than in the way one typically "expands and simplifies".
        $endgroup$
        – Omnomnomnom
        Jan 8 at 0:10










      1




      1




      $begingroup$
      This seems to be the approach being hinted at. From OP's perspective, I think the insight that was missing was that we should expand things in terms of $varphi$ and $psi$, rather than in the way one typically "expands and simplifies".
      $endgroup$
      – Omnomnomnom
      Jan 8 at 0:10






      $begingroup$
      This seems to be the approach being hinted at. From OP's perspective, I think the insight that was missing was that we should expand things in terms of $varphi$ and $psi$, rather than in the way one typically "expands and simplifies".
      $endgroup$
      – Omnomnomnom
      Jan 8 at 0:10













      7












      $begingroup$

      For what it's worth, there's a nice approach using matrices. We can characterize the recurrence by
      $$
      pmatrix{x_{n+1}\x_n} = pmatrix{1&1\1&0}pmatrix{x_{n}\x_{n-1}}, quad n geq 1
      $$
      Let $F = pmatrix{1&1\1&0}$. As a consequence of this characterization, we end up with the formula
      $$
      F^n = pmatrix{x_{n+1} & x_n\ x_n & x_{n-1}}
      $$
      We now compute
      $$
      (F^n)^2 = pmatrix{x_{n+1} & x_n\ x_n & x_{n-1}}^2 = pmatrix{x_{n+1}^2 + x_n^2 & x_{n+1}x_n + x_n x_{n-1}\
      x_{n+1}x_n + x_n x_{n-1} & x_n^2 + x_{n-1}^2}
      $$
      On the other hand,
      $$
      (F^n)^2 = F^{2n} = pmatrix{x_{2n+1} & x_{2n}\ x_{2n} & x_{2n-1}}
      $$
      Because the two matrices are equal, their upper-left entries are equal, which is exactly the desired conclusion.






      share|cite|improve this answer









      $endgroup$


















        7












        $begingroup$

        For what it's worth, there's a nice approach using matrices. We can characterize the recurrence by
        $$
        pmatrix{x_{n+1}\x_n} = pmatrix{1&1\1&0}pmatrix{x_{n}\x_{n-1}}, quad n geq 1
        $$
        Let $F = pmatrix{1&1\1&0}$. As a consequence of this characterization, we end up with the formula
        $$
        F^n = pmatrix{x_{n+1} & x_n\ x_n & x_{n-1}}
        $$
        We now compute
        $$
        (F^n)^2 = pmatrix{x_{n+1} & x_n\ x_n & x_{n-1}}^2 = pmatrix{x_{n+1}^2 + x_n^2 & x_{n+1}x_n + x_n x_{n-1}\
        x_{n+1}x_n + x_n x_{n-1} & x_n^2 + x_{n-1}^2}
        $$
        On the other hand,
        $$
        (F^n)^2 = F^{2n} = pmatrix{x_{2n+1} & x_{2n}\ x_{2n} & x_{2n-1}}
        $$
        Because the two matrices are equal, their upper-left entries are equal, which is exactly the desired conclusion.






        share|cite|improve this answer









        $endgroup$
















          7












          7








          7





          $begingroup$

          For what it's worth, there's a nice approach using matrices. We can characterize the recurrence by
          $$
          pmatrix{x_{n+1}\x_n} = pmatrix{1&1\1&0}pmatrix{x_{n}\x_{n-1}}, quad n geq 1
          $$
          Let $F = pmatrix{1&1\1&0}$. As a consequence of this characterization, we end up with the formula
          $$
          F^n = pmatrix{x_{n+1} & x_n\ x_n & x_{n-1}}
          $$
          We now compute
          $$
          (F^n)^2 = pmatrix{x_{n+1} & x_n\ x_n & x_{n-1}}^2 = pmatrix{x_{n+1}^2 + x_n^2 & x_{n+1}x_n + x_n x_{n-1}\
          x_{n+1}x_n + x_n x_{n-1} & x_n^2 + x_{n-1}^2}
          $$
          On the other hand,
          $$
          (F^n)^2 = F^{2n} = pmatrix{x_{2n+1} & x_{2n}\ x_{2n} & x_{2n-1}}
          $$
          Because the two matrices are equal, their upper-left entries are equal, which is exactly the desired conclusion.






          share|cite|improve this answer









          $endgroup$



          For what it's worth, there's a nice approach using matrices. We can characterize the recurrence by
          $$
          pmatrix{x_{n+1}\x_n} = pmatrix{1&1\1&0}pmatrix{x_{n}\x_{n-1}}, quad n geq 1
          $$
          Let $F = pmatrix{1&1\1&0}$. As a consequence of this characterization, we end up with the formula
          $$
          F^n = pmatrix{x_{n+1} & x_n\ x_n & x_{n-1}}
          $$
          We now compute
          $$
          (F^n)^2 = pmatrix{x_{n+1} & x_n\ x_n & x_{n-1}}^2 = pmatrix{x_{n+1}^2 + x_n^2 & x_{n+1}x_n + x_n x_{n-1}\
          x_{n+1}x_n + x_n x_{n-1} & x_n^2 + x_{n-1}^2}
          $$
          On the other hand,
          $$
          (F^n)^2 = F^{2n} = pmatrix{x_{2n+1} & x_{2n}\ x_{2n} & x_{2n-1}}
          $$
          Because the two matrices are equal, their upper-left entries are equal, which is exactly the desired conclusion.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 8 at 0:18









          OmnomnomnomOmnomnomnom

          128k790179




          128k790179























              0












              $begingroup$

              $x_{2n+1}=x_{2n}+x_{2n-1}=x_2x_{2n}+x_1x_{2n-1}=x_2(x_{2n-1}+x_{2n-2})+x_1x_{2n-1}=(x_1+x_2)x_{2n-1}+x_2x_{2n-2}=x_3x_{2n-1}+x_2x_{2n-2}=x_4x_{2n-2}+x_3x_{2n-3}=cdots =x_xx_n+x_{n+1}x_{n+1}=x_n^2+x_{n+1}^2.$






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                $x_{2n+1}=x_{2n}+x_{2n-1}=x_2x_{2n}+x_1x_{2n-1}=x_2(x_{2n-1}+x_{2n-2})+x_1x_{2n-1}=(x_1+x_2)x_{2n-1}+x_2x_{2n-2}=x_3x_{2n-1}+x_2x_{2n-2}=x_4x_{2n-2}+x_3x_{2n-3}=cdots =x_xx_n+x_{n+1}x_{n+1}=x_n^2+x_{n+1}^2.$






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $x_{2n+1}=x_{2n}+x_{2n-1}=x_2x_{2n}+x_1x_{2n-1}=x_2(x_{2n-1}+x_{2n-2})+x_1x_{2n-1}=(x_1+x_2)x_{2n-1}+x_2x_{2n-2}=x_3x_{2n-1}+x_2x_{2n-2}=x_4x_{2n-2}+x_3x_{2n-3}=cdots =x_xx_n+x_{n+1}x_{n+1}=x_n^2+x_{n+1}^2.$






                  share|cite|improve this answer











                  $endgroup$



                  $x_{2n+1}=x_{2n}+x_{2n-1}=x_2x_{2n}+x_1x_{2n-1}=x_2(x_{2n-1}+x_{2n-2})+x_1x_{2n-1}=(x_1+x_2)x_{2n-1}+x_2x_{2n-2}=x_3x_{2n-1}+x_2x_{2n-2}=x_4x_{2n-2}+x_3x_{2n-3}=cdots =x_xx_n+x_{n+1}x_{n+1}=x_n^2+x_{n+1}^2.$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 8 at 0:10

























                  answered Jan 7 at 23:57









                  John_WickJohn_Wick

                  1,601111




                  1,601111























                      -3












                      $begingroup$

                      Based on the title of the topic and the hint, you need to plug the Closed form expression into your equality



                      $$x_{n}^2+x_{n+1}^2-x_{2n+1}$$



                      and make sure it equals zero identically.



                      Best way is to taky any CAS and plug the figures. For Wolfram Mathematica




                      x[n_] := 1/Sqrt[5]*(((1 + Sqrt[5])/2)^n - ((1 - Sqrt[5])/2)^n);



                      x[a]^2 + x[a + 1]^2 - x[2*a + 1]



                      Simplify[Out]




                      Then you get something like this



                      $$frac{1}{5} left(left(frac{2}{sqrt{5}+1}right)^{-a-1}-left(frac{1}{2} left(1-sqrt{5}right)right)^{a+1}right)^2+frac{1}{5} left(left(frac{1}{2} left(sqrt{5}+1right)right)^a-left(frac{1}{2} left(1-sqrt{5}right)right)^aright)^2-frac{left(frac{2}{sqrt{5}+1}right)^{-2 a-1}-left(frac{1}{2} left(1-sqrt{5}right)right)^{2 a+1}}{sqrt{5}}$$



                      You need to carefully open up all brackets, cancel out the terms to get zero.






                      share|cite|improve this answer









                      $endgroup$









                      • 2




                        $begingroup$
                        relevant comic
                        $endgroup$
                        – Omnomnomnom
                        Jan 8 at 1:44
















                      -3












                      $begingroup$

                      Based on the title of the topic and the hint, you need to plug the Closed form expression into your equality



                      $$x_{n}^2+x_{n+1}^2-x_{2n+1}$$



                      and make sure it equals zero identically.



                      Best way is to taky any CAS and plug the figures. For Wolfram Mathematica




                      x[n_] := 1/Sqrt[5]*(((1 + Sqrt[5])/2)^n - ((1 - Sqrt[5])/2)^n);



                      x[a]^2 + x[a + 1]^2 - x[2*a + 1]



                      Simplify[Out]




                      Then you get something like this



                      $$frac{1}{5} left(left(frac{2}{sqrt{5}+1}right)^{-a-1}-left(frac{1}{2} left(1-sqrt{5}right)right)^{a+1}right)^2+frac{1}{5} left(left(frac{1}{2} left(sqrt{5}+1right)right)^a-left(frac{1}{2} left(1-sqrt{5}right)right)^aright)^2-frac{left(frac{2}{sqrt{5}+1}right)^{-2 a-1}-left(frac{1}{2} left(1-sqrt{5}right)right)^{2 a+1}}{sqrt{5}}$$



                      You need to carefully open up all brackets, cancel out the terms to get zero.






                      share|cite|improve this answer









                      $endgroup$









                      • 2




                        $begingroup$
                        relevant comic
                        $endgroup$
                        – Omnomnomnom
                        Jan 8 at 1:44














                      -3












                      -3








                      -3





                      $begingroup$

                      Based on the title of the topic and the hint, you need to plug the Closed form expression into your equality



                      $$x_{n}^2+x_{n+1}^2-x_{2n+1}$$



                      and make sure it equals zero identically.



                      Best way is to taky any CAS and plug the figures. For Wolfram Mathematica




                      x[n_] := 1/Sqrt[5]*(((1 + Sqrt[5])/2)^n - ((1 - Sqrt[5])/2)^n);



                      x[a]^2 + x[a + 1]^2 - x[2*a + 1]



                      Simplify[Out]




                      Then you get something like this



                      $$frac{1}{5} left(left(frac{2}{sqrt{5}+1}right)^{-a-1}-left(frac{1}{2} left(1-sqrt{5}right)right)^{a+1}right)^2+frac{1}{5} left(left(frac{1}{2} left(sqrt{5}+1right)right)^a-left(frac{1}{2} left(1-sqrt{5}right)right)^aright)^2-frac{left(frac{2}{sqrt{5}+1}right)^{-2 a-1}-left(frac{1}{2} left(1-sqrt{5}right)right)^{2 a+1}}{sqrt{5}}$$



                      You need to carefully open up all brackets, cancel out the terms to get zero.






                      share|cite|improve this answer









                      $endgroup$



                      Based on the title of the topic and the hint, you need to plug the Closed form expression into your equality



                      $$x_{n}^2+x_{n+1}^2-x_{2n+1}$$



                      and make sure it equals zero identically.



                      Best way is to taky any CAS and plug the figures. For Wolfram Mathematica




                      x[n_] := 1/Sqrt[5]*(((1 + Sqrt[5])/2)^n - ((1 - Sqrt[5])/2)^n);



                      x[a]^2 + x[a + 1]^2 - x[2*a + 1]



                      Simplify[Out]




                      Then you get something like this



                      $$frac{1}{5} left(left(frac{2}{sqrt{5}+1}right)^{-a-1}-left(frac{1}{2} left(1-sqrt{5}right)right)^{a+1}right)^2+frac{1}{5} left(left(frac{1}{2} left(sqrt{5}+1right)right)^a-left(frac{1}{2} left(1-sqrt{5}right)right)^aright)^2-frac{left(frac{2}{sqrt{5}+1}right)^{-2 a-1}-left(frac{1}{2} left(1-sqrt{5}right)right)^{2 a+1}}{sqrt{5}}$$



                      You need to carefully open up all brackets, cancel out the terms to get zero.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 8 at 0:17









                      Mikhail DMikhail D

                      34325




                      34325








                      • 2




                        $begingroup$
                        relevant comic
                        $endgroup$
                        – Omnomnomnom
                        Jan 8 at 1:44














                      • 2




                        $begingroup$
                        relevant comic
                        $endgroup$
                        – Omnomnomnom
                        Jan 8 at 1:44








                      2




                      2




                      $begingroup$
                      relevant comic
                      $endgroup$
                      – Omnomnomnom
                      Jan 8 at 1:44




                      $begingroup$
                      relevant comic
                      $endgroup$
                      – Omnomnomnom
                      Jan 8 at 1:44


















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