I would appreciate if somebody could help me with the following problem:
Q: How to proof?
If $\{a_n\}$ satisfy $a_{1}=a$, $a_2=b$, $a_{n+2}=a_{n+1}+a_{n}$($a,b$: positive integers)
then proof that $\frac{a_{4n}-a_2}{a_{2n+1}}$ : integer
I try start by mathematical induction but...., find $f(n)=\frac{a_{4n}-a_2}{a_{2n+1}}$ $f(1)=1$, $f(2)=4$, $f(3)=11$, $f(4)=29$,...
It is easy to prove by induction that $a_n=aF_{n-2}+bF_{n-1}$, where $F_k$ is Fibonacci numbers. I want to prove that $\frac{a_{4n}-a_2}{a_{2n+1}}=F_{2n-2}+F_{2n}$. Therefore, we must prove that $$aF_{4n-2}+bF_{4n-1}-b=(aF_{2n-1}+bF_{2n})(F_{2n-2}+F_{2n})$$ Therefore, we need prove that $F_{4n-2}=F_{2n-1}(F_{2n-2}+F_{2n})$ and $F_{4n-1}-1=F_{2n}(F_{2n-2}+F_{2n})$.These two equalities are easily proved by well-known formulas from Wikipedia.
Hints. Let $\phi$ and $\psi$ be the two roots of $x^2-x-1=0$. Since every Fibonacci sequence is a linear combination of the geometric progressions of $\phi$ and $\psi$, if we define $$ f_n(x)=\frac{x^{4n}-x^2}{x^{2n+1}}=x^{2n-1}-\frac{1}{x^{2n-1}}, $$ it suffices to show that $f_n(\phi)$ is an integer and $f_n(\phi)=f_n(\psi)$ for each $n$. If you want to use mathematical induction, you may need to prove, for each $n$, that $f_n(\phi)=f_n(\psi)\in\mathbb N$ and $g_n(\phi)=g_n(\psi)\in\mathbb N$, where $$ g_n(x)=x^{2n}+\frac{1}{x^{2n}}. $$
$R=\frac{a_{4n}-a_2}{a_{2n+1}}$
$a_1=a$
$a_2=b$
$a_3=a+b$
$a_4=a+2b$
$a_5=2a+3b$
$a_6=3a+5b$
$a_7=5a+8b$
$a_8=8a+13b$
For n=2 we have:
$\frac{8a+13b-b}{2a+3b}=4$
By experimental induction if R is true for n=2 it must also be true for n+1=3, we check this:
$\frac{a_{12}-b}{a_7}=\frac{55a+89b-b}{5a+8b}=11$
This true for any value of a and b,for example $a_{12}=65$ for a=0, b=1 and n=3 :
$\frac{65-1}{8}=8$
Then $F_{k+1}=AF_k$ and $F_n=A^nF_0$ with $a_0=0$.
The eigenvalues and a set of corresponding eigenvectors then give you a closed form expression of $F_n$, which means you have a closed form expression for $a_n$.
– Jun 26 '19 at 15:40