2000 Paper 3 Q8

Year: 2000
Paper: 3
Question Number: 8

Course: UFM Additional Further Pure
Section: Sequences and Series

Difficulty: 1700.0 Banger: 1484.0

sequences recurrence relation proof by induction auxiliary equation golden ratio nonlinear recurrence closed form

Problem

The sequence $a_n$ is defined by $a_0 = 1$ , $a_1 = 1$ , and $$ a_n = {1 + a_{n - 1}^2 \over a_{n - 2} } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( n \ge 2 ) . $$ Prove by induction that $$ a_n = 3 a_{n - 1} - a_{n - 2} \ \ \ \ \ \ \ \ \ \ \ ( n \ge2 ) . $$ Hence show that $$ a_n = {\alpha^{2 n - 1} + \alpha^{- ( 2 n - 1 )} \over \sqrt 5} \ \ \ \ \ \ (n\ge1), $$ where $\displaystyle{\alpha = {1 + \sqrt 5 \over 2}}$.

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Banger Rating: 1484.0

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Problem source

The sequence $a_n$ is defined by $a_0 = 1$ , $a_1 = 1$ , and 
$$ 
a_n = {1 + a_{n - 1}^2 \over a_{n - 2} } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( n \ge 2 ) . 
$$ 
Prove by induction that 
$$ 
a_n = 3 a_{n - 1} - a_{n - 2} \ \ \ \ \ \ \ \ \ \ \ ( n \ge2 ) . 
$$ 
Hence show that  
$$  
a_n = {\alpha^{2 n - 1} + \alpha^{- ( 2 n - 1 )} \over \sqrt 5} 
\ \ \ \ \ \  (n\ge1), 
$$ 
where $\displaystyle{\alpha = {1 + \sqrt 5 \over 2}}$.

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