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

[Suk-Sci]

Mathematical induction...please help me!

_{}a1 =1, _{}a2 =1,_{}a3 =2,_{}a4 =3..._{}an = _{}an-1 + _{}an-2 is a Fibonacci series...Prove That
^{2}_{}an+1 = _{}an _{}an+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]

_{}a1 =1, _{}a2 =1,_{}a3 =2,_{}a4 =3..._{}an = _{}an-1 + _{}an-2 is a Fibonacci series...Prove That
^{2}_{}an+1 = _{}an _{}an+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
a_{n+1}^2= a_n a_{n+2}+ (-1)^n[/itex]?<br /> 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.<br /> <br /> I would recommend proof by induction on n.<br /> <br /> 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}. <br /> <br /> I would now break the proof into two cases:<br /> 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}.<br /> <br /> 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...