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 2 years ago
[SOLVED] Define a sequence of real numbers by \(a_0=0\), \(a_1=1\), \(a_2=1\) and \[\large a_{n+3}=2a_{n+2}+2a_{n+1}a_n\]Show that \(a_n\) is a perfect square for every \(n\in\mathbb{N}\)
 2 years ago
[SOLVED] Define a sequence of real numbers by \(a_0=0\), \(a_1=1\), \(a_2=1\) and \[\large a_{n+3}=2a_{n+2}+2a_{n+1}a_n\]Show that \(a_n\) is a perfect square for every \(n\in\mathbb{N}\)

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bmp
 2 years ago
Best ResponseYou've already chosen the best response.0The recurrence relation seems pretty messy, so I will try to solve it inductively, T_T

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.0How about if we reduce every term into the \(ka_{n+2} + ma_{n+1} + na_n\)? The term \(a_n\) won't matter as its Zero. We can express terms as just \(k + m\). Maybe?

bmp
 2 years ago
Best ResponseYou've already chosen the best response.0I tried to find a pattern in the roots, but couldn't. a3 = 2*2, a4 = 3*3, a5 = 5*5, a6 = 8*8, a7 = 12*12. Will try something else now.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.4I would suggest induction as bmp is attempting.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.4@bmp, check your value for \(a_7\)

bmp
 2 years ago
Best ResponseYou've already chosen the best response.0I think every \[a_{n} = Fib(n)*Fib(n)\]that is a perfect square for every n in N.

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.0But how are you going to prove it?

bmp
 2 years ago
Best ResponseYou've already chosen the best response.0I am trying to refine my proof, but I think that assuming that that is true(by induction), it's not too hard to proof that it is correct.

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.0\[a_{n+3} = 2a_{n+2} + 2a_{n+1} a_n\]Lets assume \(a_{n+3} = q^2\). \[a_{n+4} = 2a_{n+3} + 2a_{n+2} a_{n+1} \implies a_{n+4} = 2q^2 + 2a_{n+2}a_{n+1}\]Now what? How do we prove \(a_{n+4}\) is perfect square as well.

bmp
 2 years ago
Best ResponseYou've already chosen the best response.0I think that the proof would go more or less like this: Assume:\[a_{k} = Fib(k)*Fib(k)\]Clearly, that holds for k <= 2. So, multiply it by 1, we get:\[a_{k} = (Fib(k)*Fib(k))\]Add 2*(ak+2 ak+1) in both sides, we get:\[2(a_{k+2} + a_{k+1})  a_{k} = 2(a_{k+2} + a_{k+1})  Fib(k)Fib(k) \rightarrow a_{k+3} = a_{k+3}\]What I just did I am pretty sure is a fallacy, but the idea is that we have to prove that every {ak} has the form Fib^2(k). I am still working on it, was never good with proofs :(

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.4I'm also pretty sure what you just did was a fallacy. However, induction is the correct path to go. Keep trying.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.4Another hint that may be helpful: Use strong induction.

bmp
 2 years ago
Best ResponseYou've already chosen the best response.0I see. It makes sense, since I will need a(1) ... a(k) to be true to prove that a(k+1) is true. This is a nice problem @KingGeorge :) Thanks for it.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.4You are also correct in that it is the sequence of squares of the Fibonacci numbers. You still need to prove it though. :)

m_charron2
 2 years ago
Best ResponseYou've already chosen the best response.1I think I may have gotten it? a5 = 2a4 + 2a3  a2 = 2 F(4) ^2 + 2 F(3) ^2  F(2) ^2 = F(4)^2 + 2F(3)^2 +(F(3)+F(2))^2  F(2)^2 = F(4)^2 + 2F(3)^2 +F(3)^2+2F(3)F(2) = F(4)^2 + 2F(3)(F(3)+F(2)) + F(3)^2 = (F(4) + F(3))^2 = (F(5)^2) and this would be true always? tadam?

bmp
 2 years ago
Best ResponseYou've already chosen the best response.0Yeah, I just now got to it. I think that would be the way, if you can prove for a(1)...a(k).

m_charron2
 2 years ago
Best ResponseYou've already chosen the best response.1well, I guess that, if a4 = F(4)^2 and a5 = F(5)^2, then a6 = F(6)^2, so a7 = F(7)^2, so a8 = F(8)^2, etc? Am I allowed to do this?

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.4Try using n, n+1, n+2, n+3 instead of 4, 5, 6, 7. Your proof above is very very close to correct. You basically just have to use a variable instead of numbers.

bmp
 2 years ago
Best ResponseYou've already chosen the best response.0@m_charron2 Kudos for beating me on the proof, mate :). Well done.

m_charron2
 2 years ago
Best ResponseYou've already chosen the best response.1oh, couldn't've done it if you hadn't told us about the fibonacci here, so Kudos to you as well I can't find how to start replacing the numbers with variables without assuming things. I can't prove that a_(n+3) = F(n+3)^2 without first knowing that a_(n+2) = F(n+2)^2, a_(n+1) = F(n+1)^2 and a_n = F(n)^2... I'm lost in a mental paradox of my own making I think...

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.4You're first method of showing it's true for \(a_5\) is so close it might as well be correct. Just substitute \[\large a_5 =a_{n+1}\]\[\large a_4 =a_{n}\]\[\large a_3 =a_{n1}\]\[\large a_2 =a_{n2}\]And similar substitutions for the Fibonacci numbers. Then go through the same steps. You will then have proved it for the general case.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.4For those interested, here's the solution I got when I was challenged with this problem.

m_charron2
 2 years ago
Best ResponseYou've already chosen the best response.1on this note, off to sleep for me, that drained me lol. Excellent problem KingGeorge!
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