Let $(F_n)_{nin mathbb{N}}$ be the Fibonacci sequence. Prove that $F_{n+1}F_n - F_{n-1}F_{n-2} = F_{2n-1}$.












5












$begingroup$


Let $(F_n)_{nin mathbb{N}}$ be the Fibonacci sequence. Prove that $F_{n+1}F_n - F_{n-1}F_{n-2} = F_{2n-1}$.



I'm trying to prove usinge the induction principle, so here is my sketch:



$(i)$ $n = 3 implies F_4F_3-F_2F_1= 6-1 = 5 = F_5$



$(ii)$ Supose true for $n = k$



$F_{k+2}F_{k+1}-F_{k}F_{k-1} = (F_{k+1}+F_k)F_{k+1} - (F_{k-1}+F_{k-2})F_{k-1} = F_{k+1}F_k-F_{k-1}F_{k-2} + F_{k+1}^2 - F_{k-1}^2 = F_{2k-1} + F_{k+1}^2 - F_{k-1}^2 $.



I got stuck here, how can I transform $F_{k+1}^2 - F_{k-1}^2$ in $F_{2k}$?






discrete-mathematics induction fibonacci-numbers







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  • 3




    $begingroup$
    Note the Fibonacci number Wikipedia entry states that a form of "d'Ocagne's identity" is $F_{2n} = F_{n + 1}^2 - F_{n - 1}^2$. Thus, doing a search on this identity may provide a relatively simple way to prove it and, thus, finish your induction proof.
    $endgroup$
    – John Omielan
    Jan 23 at 2:15


















5












$begingroup$


Let $(F_n)_{nin mathbb{N}}$ be the Fibonacci sequence. Prove that $F_{n+1}F_n - F_{n-1}F_{n-2} = F_{2n-1}$.



I'm trying to prove usinge the induction principle, so here is my sketch:



$(i)$ $n = 3 implies F_4F_3-F_2F_1= 6-1 = 5 = F_5$



$(ii)$ Supose true for $n = k$



$F_{k+2}F_{k+1}-F_{k}F_{k-1} = (F_{k+1}+F_k)F_{k+1} - (F_{k-1}+F_{k-2})F_{k-1} = F_{k+1}F_k-F_{k-1}F_{k-2} + F_{k+1}^2 - F_{k-1}^2 = F_{2k-1} + F_{k+1}^2 - F_{k-1}^2 $.



I got stuck here, how can I transform $F_{k+1}^2 - F_{k-1}^2$ in $F_{2k}$?






discrete-mathematics induction fibonacci-numbers







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Note the Fibonacci number Wikipedia entry states that a form of "d'Ocagne's identity" is $F_{2n} = F_{n + 1}^2 - F_{n - 1}^2$. Thus, doing a search on this identity may provide a relatively simple way to prove it and, thus, finish your induction proof.
    $endgroup$
    – John Omielan
    Jan 23 at 2:15
















5












5








5


2



$begingroup$


Let $(F_n)_{nin mathbb{N}}$ be the Fibonacci sequence. Prove that $F_{n+1}F_n - F_{n-1}F_{n-2} = F_{2n-1}$.



I'm trying to prove usinge the induction principle, so here is my sketch:



$(i)$ $n = 3 implies F_4F_3-F_2F_1= 6-1 = 5 = F_5$



$(ii)$ Supose true for $n = k$



$F_{k+2}F_{k+1}-F_{k}F_{k-1} = (F_{k+1}+F_k)F_{k+1} - (F_{k-1}+F_{k-2})F_{k-1} = F_{k+1}F_k-F_{k-1}F_{k-2} + F_{k+1}^2 - F_{k-1}^2 = F_{2k-1} + F_{k+1}^2 - F_{k-1}^2 $.



I got stuck here, how can I transform $F_{k+1}^2 - F_{k-1}^2$ in $F_{2k}$?






discrete-mathematics induction fibonacci-numbers







share|cite|improve this question









$endgroup$




Let $(F_n)_{nin mathbb{N}}$ be the Fibonacci sequence. Prove that $F_{n+1}F_n - F_{n-1}F_{n-2} = F_{2n-1}$.



I'm trying to prove usinge the induction principle, so here is my sketch:



$(i)$ $n = 3 implies F_4F_3-F_2F_1= 6-1 = 5 = F_5$



$(ii)$ Supose true for $n = k$



$F_{k+2}F_{k+1}-F_{k}F_{k-1} = (F_{k+1}+F_k)F_{k+1} - (F_{k-1}+F_{k-2})F_{k-1} = F_{k+1}F_k-F_{k-1}F_{k-2} + F_{k+1}^2 - F_{k-1}^2 = F_{2k-1} + F_{k+1}^2 - F_{k-1}^2 $.



I got stuck here, how can I transform $F_{k+1}^2 - F_{k-1}^2$ in $F_{2k}$?






discrete-mathematics induction fibonacci-numbers




discrete-mathematics induction fibonacci-numbers






share|cite|improve this question













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asked Jan 23 at 1:46









MrBrMrBr

756




756








  • 3




    $begingroup$
    Note the Fibonacci number Wikipedia entry states that a form of "d'Ocagne's identity" is $F_{2n} = F_{n + 1}^2 - F_{n - 1}^2$. Thus, doing a search on this identity may provide a relatively simple way to prove it and, thus, finish your induction proof.
    $endgroup$
    – John Omielan
    Jan 23 at 2:15
















  • 3




    $begingroup$
    Note the Fibonacci number Wikipedia entry states that a form of "d'Ocagne's identity" is $F_{2n} = F_{n + 1}^2 - F_{n - 1}^2$. Thus, doing a search on this identity may provide a relatively simple way to prove it and, thus, finish your induction proof.
    $endgroup$
    – John Omielan
    Jan 23 at 2:15










3




3




$begingroup$
Note the Fibonacci number Wikipedia entry states that a form of "d'Ocagne's identity" is $F_{2n} = F_{n + 1}^2 - F_{n - 1}^2$. Thus, doing a search on this identity may provide a relatively simple way to prove it and, thus, finish your induction proof.
$endgroup$
– John Omielan
Jan 23 at 2:15






$begingroup$
Note the Fibonacci number Wikipedia entry states that a form of "d'Ocagne's identity" is $F_{2n} = F_{n + 1}^2 - F_{n - 1}^2$. Thus, doing a search on this identity may provide a relatively simple way to prove it and, thus, finish your induction proof.
$endgroup$
– John Omielan
Jan 23 at 2:15












2 Answers
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First we can prove that $F_n=frac{1}{sqrt{5}}((frac{1+sqrt{5}}{2})^n-(frac{1-sqrt{5}}{2})^n)$ by induction on $n$.
Using the above formula, we can verify that $F_{n+1}F_{n}-F_{n-1}F_{n-2}=F_{2n-1}$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    I think you should use the recurrence formula of the Fibonnacci sequence, I.e
    $ F(n+2) = F(n+1) + F(n) $
    Combine this with the formula you get by induction.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
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      0












      $begingroup$

      First we can prove that $F_n=frac{1}{sqrt{5}}((frac{1+sqrt{5}}{2})^n-(frac{1-sqrt{5}}{2})^n)$ by induction on $n$.
      Using the above formula, we can verify that $F_{n+1}F_{n}-F_{n-1}F_{n-2}=F_{2n-1}$.






      share|cite|improve this answer









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        0












        $begingroup$

        First we can prove that $F_n=frac{1}{sqrt{5}}((frac{1+sqrt{5}}{2})^n-(frac{1-sqrt{5}}{2})^n)$ by induction on $n$.
        Using the above formula, we can verify that $F_{n+1}F_{n}-F_{n-1}F_{n-2}=F_{2n-1}$.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          First we can prove that $F_n=frac{1}{sqrt{5}}((frac{1+sqrt{5}}{2})^n-(frac{1-sqrt{5}}{2})^n)$ by induction on $n$.
          Using the above formula, we can verify that $F_{n+1}F_{n}-F_{n-1}F_{n-2}=F_{2n-1}$.






          share|cite|improve this answer









          $endgroup$



          First we can prove that $F_n=frac{1}{sqrt{5}}((frac{1+sqrt{5}}{2})^n-(frac{1-sqrt{5}}{2})^n)$ by induction on $n$.
          Using the above formula, we can verify that $F_{n+1}F_{n}-F_{n-1}F_{n-2}=F_{2n-1}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 23 at 2:44









          NotoriousJuanGNotoriousJuanG

          843




          843























              0












              $begingroup$

              I think you should use the recurrence formula of the Fibonnacci sequence, I.e
              $ F(n+2) = F(n+1) + F(n) $
              Combine this with the formula you get by induction.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                I think you should use the recurrence formula of the Fibonnacci sequence, I.e
                $ F(n+2) = F(n+1) + F(n) $
                Combine this with the formula you get by induction.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I think you should use the recurrence formula of the Fibonnacci sequence, I.e
                  $ F(n+2) = F(n+1) + F(n) $
                  Combine this with the formula you get by induction.






                  share|cite|improve this answer









                  $endgroup$



                  I think you should use the recurrence formula of the Fibonnacci sequence, I.e
                  $ F(n+2) = F(n+1) + F(n) $
                  Combine this with the formula you get by induction.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 23 at 8:20









                  Cauchy is my masterCauchy is my master

                  413




                  413






























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