Show that $F_{3n} = F_{n}(L_{2n} + (-1)^n)$












1












$begingroup$


Let $F_n, L_n$ be the Fibonacci and Lucas sequences respectively.
Show that $F_{3n} = F_{n}(L_{2n} + (-1)^n)$. In my attempt I am using Binet's formula, and the equivalent for the Lucas numbers.



begin{align}
F_{3n} &= frac {alpha^{3n} - beta^{3n}} { sqrt{5}} = F_{n}(L_{2n} + (-1)^n) \ &= frac {alpha^{n} - beta^{n}} {sqrt{5}}(alpha^{2n}+beta^{2n}+(-1)^n) \
&=frac {alpha^{3n} + alpha^nbeta^{2n}+(-1)^nalpha^n-beta^nalpha^{2n}-beta^{3n}-(-1)^nbeta^n}{sqrt{5}},
end{align}
from here I am not sure how to proceed. Hints appreciated.






fibonacci-numbers lucas-numbers







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$endgroup$












  • $begingroup$
    I really don't think your equation is right. Maybe $F(3n)$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 16 at 17:54










  • $begingroup$
    @LordSharktheUnknown sorry, edited.
    $endgroup$
    – IntegrateThis
    Jan 16 at 17:54






  • 1




    $begingroup$
    Recall that $alphabeta=-1$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 16 at 17:59










  • $begingroup$
    @LordSharktheUnknown ok that's what I needed thanks.
    $endgroup$
    – IntegrateThis
    Jan 16 at 18:33
















1












$begingroup$


Let $F_n, L_n$ be the Fibonacci and Lucas sequences respectively.
Show that $F_{3n} = F_{n}(L_{2n} + (-1)^n)$. In my attempt I am using Binet's formula, and the equivalent for the Lucas numbers.



begin{align}
F_{3n} &= frac {alpha^{3n} - beta^{3n}} { sqrt{5}} = F_{n}(L_{2n} + (-1)^n) \ &= frac {alpha^{n} - beta^{n}} {sqrt{5}}(alpha^{2n}+beta^{2n}+(-1)^n) \
&=frac {alpha^{3n} + alpha^nbeta^{2n}+(-1)^nalpha^n-beta^nalpha^{2n}-beta^{3n}-(-1)^nbeta^n}{sqrt{5}},
end{align}
from here I am not sure how to proceed. Hints appreciated.






fibonacci-numbers lucas-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    I really don't think your equation is right. Maybe $F(3n)$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 16 at 17:54










  • $begingroup$
    @LordSharktheUnknown sorry, edited.
    $endgroup$
    – IntegrateThis
    Jan 16 at 17:54






  • 1




    $begingroup$
    Recall that $alphabeta=-1$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 16 at 17:59










  • $begingroup$
    @LordSharktheUnknown ok that's what I needed thanks.
    $endgroup$
    – IntegrateThis
    Jan 16 at 18:33














1












1








1


1



$begingroup$


Let $F_n, L_n$ be the Fibonacci and Lucas sequences respectively.
Show that $F_{3n} = F_{n}(L_{2n} + (-1)^n)$. In my attempt I am using Binet's formula, and the equivalent for the Lucas numbers.



begin{align}
F_{3n} &= frac {alpha^{3n} - beta^{3n}} { sqrt{5}} = F_{n}(L_{2n} + (-1)^n) \ &= frac {alpha^{n} - beta^{n}} {sqrt{5}}(alpha^{2n}+beta^{2n}+(-1)^n) \
&=frac {alpha^{3n} + alpha^nbeta^{2n}+(-1)^nalpha^n-beta^nalpha^{2n}-beta^{3n}-(-1)^nbeta^n}{sqrt{5}},
end{align}
from here I am not sure how to proceed. Hints appreciated.






fibonacci-numbers lucas-numbers







share|cite|improve this question











$endgroup$




Let $F_n, L_n$ be the Fibonacci and Lucas sequences respectively.
Show that $F_{3n} = F_{n}(L_{2n} + (-1)^n)$. In my attempt I am using Binet's formula, and the equivalent for the Lucas numbers.



begin{align}
F_{3n} &= frac {alpha^{3n} - beta^{3n}} { sqrt{5}} = F_{n}(L_{2n} + (-1)^n) \ &= frac {alpha^{n} - beta^{n}} {sqrt{5}}(alpha^{2n}+beta^{2n}+(-1)^n) \
&=frac {alpha^{3n} + alpha^nbeta^{2n}+(-1)^nalpha^n-beta^nalpha^{2n}-beta^{3n}-(-1)^nbeta^n}{sqrt{5}},
end{align}
from here I am not sure how to proceed. Hints appreciated.






fibonacci-numbers lucas-numbers




fibonacci-numbers lucas-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









Leucippus

19.8k102871




19.8k102871










asked Jan 16 at 17:51









IntegrateThisIntegrateThis

1,9611818




1,9611818












  • $begingroup$
    I really don't think your equation is right. Maybe $F(3n)$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 16 at 17:54










  • $begingroup$
    @LordSharktheUnknown sorry, edited.
    $endgroup$
    – IntegrateThis
    Jan 16 at 17:54






  • 1




    $begingroup$
    Recall that $alphabeta=-1$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 16 at 17:59










  • $begingroup$
    @LordSharktheUnknown ok that's what I needed thanks.
    $endgroup$
    – IntegrateThis
    Jan 16 at 18:33


















  • $begingroup$
    I really don't think your equation is right. Maybe $F(3n)$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 16 at 17:54










  • $begingroup$
    @LordSharktheUnknown sorry, edited.
    $endgroup$
    – IntegrateThis
    Jan 16 at 17:54






  • 1




    $begingroup$
    Recall that $alphabeta=-1$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 16 at 17:59










  • $begingroup$
    @LordSharktheUnknown ok that's what I needed thanks.
    $endgroup$
    – IntegrateThis
    Jan 16 at 18:33
















$begingroup$
I really don't think your equation is right. Maybe $F(3n)$?
$endgroup$
– Lord Shark the Unknown
Jan 16 at 17:54




$begingroup$
I really don't think your equation is right. Maybe $F(3n)$?
$endgroup$
– Lord Shark the Unknown
Jan 16 at 17:54












$begingroup$
@LordSharktheUnknown sorry, edited.
$endgroup$
– IntegrateThis
Jan 16 at 17:54




$begingroup$
@LordSharktheUnknown sorry, edited.
$endgroup$
– IntegrateThis
Jan 16 at 17:54




1




1




$begingroup$
Recall that $alphabeta=-1$.
$endgroup$
– Lord Shark the Unknown
Jan 16 at 17:59




$begingroup$
Recall that $alphabeta=-1$.
$endgroup$
– Lord Shark the Unknown
Jan 16 at 17:59












$begingroup$
@LordSharktheUnknown ok that's what I needed thanks.
$endgroup$
– IntegrateThis
Jan 16 at 18:33




$begingroup$
@LordSharktheUnknown ok that's what I needed thanks.
$endgroup$
– IntegrateThis
Jan 16 at 18:33










1 Answer
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$begingroup$

Starting with $F_{3n}$ then
begin{align}
F_{3n} &= frac{1}{sqrt{5}} , ( alpha^{3n} - beta^{3n} ) \
&= frac{1}{sqrt{5}} , (alpha^{3n} - 3 , alpha^{2n} beta^n + 3 , alpha^n beta^{2n} - beta^{3n}) + frac{3 (-1)^n}{sqrt{5}} , (alpha^n - beta^n) \
&= frac{1}{sqrt{5}} , (alpha^n - beta^n)^3 + 3 , (-1)^n , F_{n} \
&= F_{n} , (5 F_{n}^2 + 3(-1)^n) \
&= F_{n} , ( (alpha^n - beta^n)^2 + 3 (-1)^n) = F_{n} , (alpha^{2n} + beta^{2n} + (-1)^n) \
&= F_{n} , (L_{2n} + (-1)^n).
end{align}






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    $begingroup$

    Starting with $F_{3n}$ then
    begin{align}
    F_{3n} &= frac{1}{sqrt{5}} , ( alpha^{3n} - beta^{3n} ) \
    &= frac{1}{sqrt{5}} , (alpha^{3n} - 3 , alpha^{2n} beta^n + 3 , alpha^n beta^{2n} - beta^{3n}) + frac{3 (-1)^n}{sqrt{5}} , (alpha^n - beta^n) \
    &= frac{1}{sqrt{5}} , (alpha^n - beta^n)^3 + 3 , (-1)^n , F_{n} \
    &= F_{n} , (5 F_{n}^2 + 3(-1)^n) \
    &= F_{n} , ( (alpha^n - beta^n)^2 + 3 (-1)^n) = F_{n} , (alpha^{2n} + beta^{2n} + (-1)^n) \
    &= F_{n} , (L_{2n} + (-1)^n).
    end{align}






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Starting with $F_{3n}$ then
      begin{align}
      F_{3n} &= frac{1}{sqrt{5}} , ( alpha^{3n} - beta^{3n} ) \
      &= frac{1}{sqrt{5}} , (alpha^{3n} - 3 , alpha^{2n} beta^n + 3 , alpha^n beta^{2n} - beta^{3n}) + frac{3 (-1)^n}{sqrt{5}} , (alpha^n - beta^n) \
      &= frac{1}{sqrt{5}} , (alpha^n - beta^n)^3 + 3 , (-1)^n , F_{n} \
      &= F_{n} , (5 F_{n}^2 + 3(-1)^n) \
      &= F_{n} , ( (alpha^n - beta^n)^2 + 3 (-1)^n) = F_{n} , (alpha^{2n} + beta^{2n} + (-1)^n) \
      &= F_{n} , (L_{2n} + (-1)^n).
      end{align}






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Starting with $F_{3n}$ then
        begin{align}
        F_{3n} &= frac{1}{sqrt{5}} , ( alpha^{3n} - beta^{3n} ) \
        &= frac{1}{sqrt{5}} , (alpha^{3n} - 3 , alpha^{2n} beta^n + 3 , alpha^n beta^{2n} - beta^{3n}) + frac{3 (-1)^n}{sqrt{5}} , (alpha^n - beta^n) \
        &= frac{1}{sqrt{5}} , (alpha^n - beta^n)^3 + 3 , (-1)^n , F_{n} \
        &= F_{n} , (5 F_{n}^2 + 3(-1)^n) \
        &= F_{n} , ( (alpha^n - beta^n)^2 + 3 (-1)^n) = F_{n} , (alpha^{2n} + beta^{2n} + (-1)^n) \
        &= F_{n} , (L_{2n} + (-1)^n).
        end{align}






        share|cite|improve this answer









        $endgroup$



        Starting with $F_{3n}$ then
        begin{align}
        F_{3n} &= frac{1}{sqrt{5}} , ( alpha^{3n} - beta^{3n} ) \
        &= frac{1}{sqrt{5}} , (alpha^{3n} - 3 , alpha^{2n} beta^n + 3 , alpha^n beta^{2n} - beta^{3n}) + frac{3 (-1)^n}{sqrt{5}} , (alpha^n - beta^n) \
        &= frac{1}{sqrt{5}} , (alpha^n - beta^n)^3 + 3 , (-1)^n , F_{n} \
        &= F_{n} , (5 F_{n}^2 + 3(-1)^n) \
        &= F_{n} , ( (alpha^n - beta^n)^2 + 3 (-1)^n) = F_{n} , (alpha^{2n} + beta^{2n} + (-1)^n) \
        &= F_{n} , (L_{2n} + (-1)^n).
        end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        LeucippusLeucippus

        19.8k102871




        19.8k102871






























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