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### Dhaka Secondary 2009/5

Posted: 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 $a_{2009}$.

### Re: Dhaka Secondary 2009/5

Posted: Tue Feb 01, 2011 11:42 pm
Hint:

### Re: Dhaka Secondary 2009/5

Posted: Fri Jan 06, 2012 9:02 pm

### Re: Dhaka Secondary 2009/5

Posted: Fri Jan 06, 2012 9:26 pm
How do you make a recursion ??? I can't understand ?

### Re: Dhaka Secondary 2009/5

Posted: Fri Jan 06, 2012 9:33 pm
just go some steps calculating.it'll help.
@shabnoor vai

### Re: Dhaka Secondary 2009/5

Posted: 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.

### Re: Dhaka Secondary 2009/5

Posted: 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, $a_{n}(a_n-2)+2=2+a_0 \cdot a_1 \cdot a_2 \cdot a_3 \cdots a_{n-1} \cdot a_n = a_{n+1}$

### Re: Dhaka Secondary 2009/5

Posted: Sat Jan 07, 2012 11:55 am
Wow, good one from mahi. Thanks

### Re: Dhaka Secondary 2009/5

Posted: 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?

### Re: Dhaka Secondary 2009/5

Posted: 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 $a_n=2^{2^n}+1$