Prove L_n = L_n-1 + L_n-2 for n >= 3, where L_n are Lucas numbers.
Thank you!
Hello, yc6489!
Sorry, I don't understand the question . . .
Prove $\displaystyle L_n \:= \:L_{n-1} + L_{n-2}$ for $\displaystyle n \geq 3$, where $\displaystyle L_n$ are Lucas numbers.
But that is the definition of Lucas Numbers.
. . $\displaystyle L_n\:=\:L_{n-1} + L_{n-2}$ where $\displaystyle L_1 = 1,\:L_2 = 2$
So what is there to prove?
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It's similar to saying:
"Prove that $\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots$ is the Harmonic Series."
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Fibonacci started with 1 and 1.
Lucas started with 1 and 2 ... and gets into the history books.
Okay . . . $\displaystyle S_1 = 1,\:S_2 = 4$
. . . . . . .$\displaystyle S_n\:=\:S_{n-1} + S_{n-2}$ for $\displaystyle n \geq 3.$
We have the sequence: .$\displaystyle 1,\,4,\,5,\,9,\,14,\,23,\,\hdots$
. . which are the lesser-known Soroban numbers.
What you are asked to prove is the usual definition of the Lucas numbers (when you add in
the initial valuse L_0=2, L_1=1, but we can use an alternative definition and prove the
usual one from it.
Take as the definition of Lucas number L_n, n>=1, (and without restiction if we extend the
Fibonacci numbers to negative index):
L_n=F_(n-1)+F_(n+1),
then if n>=3:
L_n=[F_(n-3)+F_(n-2)]+[F_(n-1)+F_(n)]
.....=[F_(n-3)+F_(n-1)]+[F_(n-2)+F_(n)]
.....=L_(n-2)+L_(n-1)
RonL
I think he acomplished something, they would just not place his name without any good reason.*Fibonacci started with 1 and 1.
Lucas started with 1 and 2 ... and gets into the history books.
*)But it has happened. The Pell equation is named after someone who did nothing with it, rather then the actual inventor, Fermat.
*)The L'Hopital rule is named after Guillalame de L'Hopital (I wish my name was that) because he wrote the first Calculus book ever and it featured it. While the actual discoverer was Johann Bernouilli.