Dhaka Secondary 2009/5
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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
Hint:
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
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- Nadim Ul Abrar
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Re: Dhaka Secondary 2009/5
How do you make a recursion ??? I can't understand ?
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- nafistiham
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Re: Dhaka Secondary 2009/5
just go some steps calculating.it'll help.
@shabnoor vai
@shabnoor vai
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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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.
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.........
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
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, $a_{n}(a_n-2)+2=2+a_0 \cdot a_1 \cdot a_2 \cdot a_3 \cdots a_{n-1} \cdot a_n = a_{n+1}$
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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Nur Muhammad Shafiullah | Mahi
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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.........
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
Re: Dhaka Secondary 2009/5
but how can we get $a_{2009}$ from this equation?*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, $a_{n}(a_n-2)+2=2+a_0 \cdot a_1 \cdot a_2 \cdot a_3 \cdots a_{n-1} \cdot a_n = a_{n+1}$
অম্লান সাহা
Re: Dhaka Secondary 2009/5
From the last equation, $(a_n-1)^2+1=a_{n+1}$amlansaha wrote:but how can we get $a_{2009}$ from this equation?*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, $a_{n}(a_n-2)+2=2+a_0 \cdot a_1 \cdot a_2 \cdot a_3 \cdots a_{n-1} \cdot a_n = a_{n+1}$
So $a_n=2^{2^n}+1$
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