## Dhaka Secondary 2009/5

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
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Please don't post problems (by starting a topic) in the "Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
BdMO
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### Dhaka Secondary 2009/5

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 $a_{2009}$.

Moon
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### Re: Dhaka Secondary 2009/5

Hint:
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

learn how to write equations, and don't forget to read Forum Guide and Rules.

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### Re: Dhaka Secondary 2009/5

$\frac{1}{0}$

sm.joty
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### Re: Dhaka Secondary 2009/5

How do you make a recursion ??? I can't understand ?
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........

nafistiham
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### Re: Dhaka Secondary 2009/5

just go some steps calculating.it'll help.
@shabnoor vai
$\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.

sm.joty
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### Re: Dhaka Secondary 2009/5

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.
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........

*Mahi*
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### Re: Dhaka Secondary 2009/5

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

Use $L^AT_EX$, It makes our work a lot easier!

sm.joty
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### Re: Dhaka Secondary 2009/5

Wow, good one from mahi. Thanks
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........

amlansaha
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### Re: Dhaka Secondary 2009/5

*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

*Mahi*
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### Re: Dhaka Secondary 2009/5

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 $a_n=2^{2^n}+1$
Use $L^AT_EX$, It makes our work a lot easier!