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 February 25th, 2015, 05:54 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, 06:11 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms $$56  February 25th, 2015, 07:12 PM #3 Math Team Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra I get the other value. A Perl script: Code: my @a = (1, 1); while ( @a < 2010 ) { push( @a, a[-2] + 1 / a[-1] ); printf STDOUT "%s: %4.3f\n", 'a_' . scalar( @a ), a[-1]; }  February 25th, 2015, 07:57 PM #4 Math Team Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra$$\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, 02:01 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, 03:44 AM   #6
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Quote:
 Originally Posted by nandofab thank u guys. The answer is: $\prod_{j=1}^{2009}(1 + \frac{1}{j+1})$
That's more or less the answer I gave: $$1 + {1 \over j+1} = {j+2 \over j+1}$$
They do seem to have an extra term though.

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