Dhaka Secondary 2009/5

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BdMO
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Dhaka Secondary 2009/5

Unread post by BdMO » Fri Jan 21, 2011 6:15 pm

Sequence $(a_n) \; ( n \geq 0 )$is defined recursively by $a_0=3$, $a_n=2+a_0 \cdot a_1 \cdot a_2 \cdot a_3 \cdots a_{n-1}, n\geq 1$. Determine $a_{2009}$.

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Moon
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Re: Dhaka Secondary 2009/5

Unread post by Moon » Tue Feb 01, 2011 11:42 pm

Hint:
Heard of Fermat's Numbers? \[F_n=2^{2^n}\]
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Nadim Ul Abrar
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Re: Dhaka Secondary 2009/5

Unread post by Nadim Ul Abrar » Fri Jan 06, 2012 9:02 pm

$a_n=2^{2^{n}}+1$
so $a_{2009} =2^{2^{2009}}+1$
$\frac{1}{0}$

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sm.joty
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Re: Dhaka Secondary 2009/5

Unread post by sm.joty » Fri Jan 06, 2012 9:26 pm

How do you make a recursion ??? I can't understand ? :(
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Re: Dhaka Secondary 2009/5

Unread post by nafistiham » Fri Jan 06, 2012 9:33 pm

just go some steps calculating.it'll help. ;)
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sm.joty
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Re: Dhaka Secondary 2009/5

Unread post by sm.joty » Sat Jan 07, 2012 1:50 am

I do it by checking and making a conjecture. Then use induction. :?
But is there any generalized way to establish a recursion ?
May be I need to study some about it. :ugeek:
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Re: Dhaka Secondary 2009/5

Unread post by *Mahi* » Sat Jan 07, 2012 11:35 am

$a_n=2+a_0 \cdot a_1 \cdot a_2 \cdot a_3 \cdots a_{n-1}$ So $a_n-2=a_0 \cdot a_1 \cdot a_2 \cdot a_3 \cdots a_{n-1}$
Or, $a_{n}(a_n-2)+2=2+a_0 \cdot a_1 \cdot a_2 \cdot a_3 \cdots a_{n-1} \cdot a_n = a_{n+1}$
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sm.joty
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Re: Dhaka Secondary 2009/5

Unread post by sm.joty » Sat Jan 07, 2012 11:55 am

Wow, good one from mahi. Thanks :mrgreen:
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Re: Dhaka Secondary 2009/5

Unread post by amlansaha » Sat Jan 07, 2012 10:00 pm

*Mahi* wrote:$a_n=2+a_0 \cdot a_1 \cdot a_2 \cdot a_3 \cdots a_{n-1}$ So $a_n-2=a_0 \cdot a_1 \cdot a_2 \cdot a_3 \cdots a_{n-1}$
Or, $a_{n}(a_n-2)+2=2+a_0 \cdot a_1 \cdot a_2 \cdot a_3 \cdots a_{n-1} \cdot a_n = a_{n+1}$
but how can we get $a_{2009}$ from this equation?
অম্লান সাহা

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*Mahi*
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Re: Dhaka Secondary 2009/5

Unread post by *Mahi* » Sat Jan 07, 2012 11:05 pm

amlansaha wrote:
*Mahi* wrote:$a_n=2+a_0 \cdot a_1 \cdot a_2 \cdot a_3 \cdots a_{n-1}$ So $a_n-2=a_0 \cdot a_1 \cdot a_2 \cdot a_3 \cdots a_{n-1}$
Or, $a_{n}(a_n-2)+2=2+a_0 \cdot a_1 \cdot a_2 \cdot a_3 \cdots a_{n-1} \cdot a_n = a_{n+1}$
but how can we get $a_{2009}$ from this equation?
From the last equation, $(a_n-1)^2+1=a_{n+1}$
So $a_n=2^{2^n}+1$
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