For integers n ≥ 2 , evaluate F n 4 − F n − 2 F n − 1 F n + 1 F n + 2 . Notation
F n denotes the n th Fibonacci number .
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Nice, you proved Cassini's identity!
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Oh lol I had never heard of that before. I don't really know much about Fibonacci identities so I wanted to prove it from scratch. It was fun!
The problem is the Gelin-Cesàro identity as mentioned by Rohit Udaiwal .
F n 4 − F n − 2 F n − 1 F n + 1 F n + 2 = 1
It can be proven using Catalan's identity which is as follows:
F n 2 − F n − r F n + r ⇒ F n − r F n + r = ( − 1 ) n − r F r 2 = F n 2 − ( − 1 ) n − r F r 2
Therefore,
F n 4 − F n − 2 F n − 1 F n + 1 F n + 2 = F n 4 − F n − 2 F n + 2 F n − 1 F n + 1 = F n 4 − ( F n 2 − ( − 1 ) n − 2 F 2 2 ) ( F n 2 − ( − 1 ) n − 1 F 1 2 ) = F n 4 − ( F n 2 ± 1 ) ( F n 2 ∓ 1 ) = F n 4 − ( F n 4 − 1 ) = 1
This identity is called Gelin-Cesàro Identity.
Like a cheater, we can use2,3,5,8,13,21,34 to solve the problem. Now putting into the problem we get=8^4-(3×5×13×21)=1. As problem states its a general constant so answer is=1.
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Firstly, we make two claims, which we use induction to prove.
Claim 1: F n 2 − F n − 1 F n + 1 = ( − 1 ) n − 1 for n ≥ 2 .
The base case n = 2 gives 1 2 − ( 1 ) ( 2 ) = − 1 , which is certainly true.
Next, assume the claim holds for a certain n = k .
When n = k + 1 ,
F k + 1 2 − F k F k + 2
= F k + 1 2 − F k ( F k + F k + 1 )
= F k + 1 2 − F k 2 − F k F k + 1
= F k + 1 2 − ( F k − 1 F k + 1 + ( − 1 ) k − 1 ) − F k F k + 1
(Using F k 2 = F k − 1 F k + 1 + ( − 1 ) k − 1 from the induction hypothesis)
= F k + 1 2 − F k − 1 F k + 1 − F k F k + 1 + ( − 1 ) k
= F k + 1 2 − ( F k − 1 + F k ) F k + 1 + ( − 1 ) k
= F k + 1 2 − F k + 1 2 + ( − 1 ) k
= ( − 1 ) k
Hence if the claim holds for n = k , then we have shown that it holds for n = k + 1 , completing the induction.
Claim 2: F n 2 − F n − 2 F n + 2 = ( − 1 ) n for n ≥ 3 .
The base case n = 3 gives 2 2 − ( 1 ) ( 5 ) = − 1 , which is certainly true.
Next, assume the claim holds for a certain n = k . To make induction easier, we rewrite the equation in terms of F n − 2 and F n by noting that F n + 2 = 3 F n − F n − 2 .
The equivalent induction hypothesis is then F k 2 + F k − 2 2 − 3 F k F k − 2 = ( − 1 ) k .
Also, note that F k + 1 = 2 F k − F k − 2 , F k − 1 = F k − F k − 2 and F k + 3 = 5 F k − 2 F k − 2 .
When n = k + 1 ,
F k + 1 2 − F k − 1 F k + 3
= ( 2 F k − F k − 2 ) 2 − ( F k − F k − 2 ) ( 5 F k − 2 F k − 2 )
= 4 F k 2 − 4 F k F k − 2 + F k − 2 2 − 5 F k 2 + 2 F k F k − 2 + 5 F k F k − 2 − 2 F k 2
= − F k 2 + 3 F k F k − 2 − F k − 2 2
= − ( F k 2 + F k − 2 2 − 3 F k F k − 2 )
= ( − 1 ) k + 1
Hence if the claim holds for n = k , then we have shown that it holds for n = k + 1 , completing the induction.
Now, we can compute the expression in the question:
F n 4 − F n − 2 F n − 1 F n + 1 F n + 2
= F n 4 − ( F n − 2 F n + 2 ) ( F n − 1 F n + 1 )
= F n 4 − ( F n 2 − ( − 1 ) n ) ( F n 2 + ( − 1 ) n )
= F n 4 − F n 4 + ( − 1 ) 2 n
= 1 for n ≥ 3 .