Can you Prove that {2}_{}an+1 = _{}an _{}an+2 =(-1)n for the Fibonacci series?

[Suk-Sci]

Mathematical induction...please help me!

$$_{}a$$1 =1, $$_{}a$$2 =1,$$_{}a$$3 =2,$$_{}a$$4 =3...$$_{}a$$n = $$_{}a$$n-1 + $$_{}a$$n-2 is a Fibonacci series...Prove That
$$^{2}_{}a$$n+1 = $$_{}a$$n $$_{}a$$n+2 =(-1)ⁿ

 

[Suk-Sci]

that is a ^2_n+1

 

[Borek]

Sorry, but your mix of LaTeX and forum formatting is completely unreadable. Check out this thread: https://www.physicsforums.com/showthread.php?t=386951 and try to repost the equation.

 

[HallsofIvy]

$$_{}a$$1 =1, $$_{}a$$2 =1,$$_{}a$$3 =2,$$_{}a$$4 =3...$$_{}a$$n = $$_{}a$$n-1 + $$_{}a$$n-2 is a Fibonacci series...Prove That
$$^{2}_{}a$$n+1 = $$_{}a$$n $$_{}a$$n+2 =(-1)ⁿ

It is far better to put "[ tex ]" or "[ itex ] [/itex ]" tagas around entire equations rather than bits and pieces!

$a_1= 1$, $a_2= 1$, $a_{n+2}= a_{n+1}+ a_n$
is a Fibonacci series.

I believe you also have an "=" where should have a "+". I think you want to prove that
[tex]a_{n+1}^2= a_n a_{n+2}+ (-1)^n[/itex]?
In the case that n=1, for example, $a_1= 1$, $a_2= 1$, and $a_3= 2$ so your formula becomes $1^2= 1(2)+ (-1)$ which is true. If n= 2, $a_2= 1$, $a_3= 2$, and $a_ 3$ so your formula becomes $2^2= (1)(3)+ 1$ which is true.

I would recommend proof by induction on n.

Suppose $a_{k+1}^2= a_k a_{k+2}+ (-1)^k$. You now want to prove that $a_{k+2}^2= a_{k+1}a_{k+3}+ (-1)^{k+1}$.

I would now break the proof into two cases:
1) k is odd. You have that $a_{k+1}^2= a_k a_{k+2}- 1$ and want to prove that $a_{k+2}^2= a_{k+1}a_{k+3}+ 1$ where $a_{k+3}= a_{k+1}+ a_{k+2}$ so that $a_{k+2}= a_{k+3}- a_{k+1}$.

2) k is even. You have that $a_{k+1}^2= a_k a_{k+2}+ 1$ and want to prove that $a_{k+2}^2= a_{k+1}a_{k+3}- 1$ where $a_{k+3}= a_{k+1}+ a_{k+2}$ so that $a_{k+2}= a_{k+3}- a_{k+1}$.

 

[Suk-Sci]

Thankx...