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February 25th, 2015, 06:54 PM   #1
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Sequence

The sequence \[a_{n}\] satisfies a1=a2=1 and \[a_{n+2} = \frac{1}{a_{n+1}} + a_{n}\] for all n. Find \[a_{2010}\]
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February 25th, 2015, 07:11 PM   #2
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$$
56<a_{2010}<57.
$$

What tools are you allowed to use?
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February 25th, 2015, 08:12 PM   #3
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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];
}
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February 25th, 2015, 08:57 PM   #4
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$$\begin{aligned} && a_{n+2} &= \frac{1}{a_{n+1}} + a_{n} \\ && a_{n+2}a_{n+1} - a_{n+1}a_{n} &= 1 \\ &\text{write $b_n = a_{n+1}a_n$, then} & b_{n+1} - b_n &= 1 \\ &\text{we also have} & a_1 = a_2 = 1 \implies b_1 &= 1 \\ &\text{so} & b_n &= n \\[12pt] &\text{now} & a_{2010} = a_{2010}a_1 &= {a_{2010}a_{2009} \cdot a_{2008}a_{2007} \cdots a_{2}a_{1} \over a_{2009} a_{2008} \cdot a_{2007}a_{2006} \cdots a_{3}a_{2} } = { b_{2009} b_{2007} \cdots b_1 \over b_{2008} b_{2006} \cdots b_2 } \\ && &= { 2009 \cdot 2007 \cdots 1 \over 2008 \cdot 2006 \cdots 2 } \end{aligned}$$
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February 26th, 2015, 03:01 AM   #5
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thank u guys. The answer is: \[\prod_{j=1}^{2009}(1 + \frac{1}{j+1})\]
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February 26th, 2015, 04:44 AM   #6
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Quote:
Originally Posted by nandofab View Post
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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