February 25th, 2015, 06:54 PM  #1 
Newbie Joined: Feb 2015 From: Brazil Posts: 23 Thanks: 0  Sequence
The sequence \[a_{n}\] satisfies a1=a2=1 and \[a_{n+2} = \frac{1}{a_{n+1}} + a_{n}\] for all n. Find \[a_{2010}\]

February 25th, 2015, 07:11 PM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 932 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
$$ 56<a_{2010}<57. $$ What tools are you allowed to use? 
February 25th, 2015, 08:12 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,854 Thanks: 2228 Math Focus: Mainly analysis and algebra 
I get the other value. A Perl script: Code: my @a = (1, 1); while ( @a < 2010 ) { push( @a, $a[2] + 1 / $a[1] ); printf STDOUT "%s: %4.3f\n", 'a_' . scalar( @a ), $a[1]; } 
February 25th, 2015, 08:57 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,854 Thanks: 2228 Math Focus: Mainly analysis and algebra 
$$\begin{aligned} && a_{n+2} &= \frac{1}{a_{n+1}} + a_{n} \\ && a_{n+2}a_{n+1}  a_{n+1}a_{n} &= 1 \\ &\text{write $b_n = a_{n+1}a_n$, then} & b_{n+1}  b_n &= 1 \\ &\text{we also have} & a_1 = a_2 = 1 \implies b_1 &= 1 \\ &\text{so} & b_n &= n \\[12pt] &\text{now} & a_{2010} = a_{2010}a_1 &= {a_{2010}a_{2009} \cdot a_{2008}a_{2007} \cdots a_{2}a_{1} \over a_{2009} a_{2008} \cdot a_{2007}a_{2006} \cdots a_{3}a_{2} } = { b_{2009} b_{2007} \cdots b_1 \over b_{2008} b_{2006} \cdots b_2 } \\ && &= { 2009 \cdot 2007 \cdots 1 \over 2008 \cdot 2006 \cdots 2 } \end{aligned}$$

February 26th, 2015, 03:01 AM  #5 
Newbie Joined: Feb 2015 From: Brazil Posts: 23 Thanks: 0 
thank u guys. The answer is: \[\prod_{j=1}^{2009}(1 + \frac{1}{j+1})\]

February 26th, 2015, 04:44 AM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,854 Thanks: 2228 Math Focus: Mainly analysis and algebra  

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