AN INTERESTING PROBLEM BY SAKAL DA

For discussing Olympiad Level Algebra (and Inequality) problems
MATHPRITOM
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AN INTERESTING PROBLEM BY SAKAL DA

Unread post by MATHPRITOM » Sun Feb 19, 2012 10:59 am

$ {a_0}=2,$

$ {a_n}=2012+{a_0}{a_1}{a_2}{a_3}...{a_{n-2}}{a_{n-1}} $ .

Find out $ {a_{2012}} $.

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SANZEED
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Re: AN INTERESTING PROBLEM BY SAKAL DA

Unread post by SANZEED » Thu Mar 01, 2012 11:24 pm

The answer is:
$a_{2012}=2012+2^{2^{2011}}$ :D
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

MATHPRITOM
Posts: 190
Joined: Sat Apr 23, 2011 8:55 am
Location: Khulna

Re: AN INTERESTING PROBLEM BY SAKAL DA

Unread post by MATHPRITOM » Fri Mar 02, 2012 1:49 am

Hi,little brother ,plz, give the full solution.

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SANZEED
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Re: AN INTERESTING PROBLEM BY SAKAL DA

Unread post by SANZEED » Sun Mar 04, 2012 4:15 pm

upssssss........sorrry..........got the mistake probably :oops: :oops: :oops:
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

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zadid xcalibured
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Re: AN INTERESTING PROBLEM BY SAKAL DA

Unread post by zadid xcalibured » Thu Feb 28, 2013 3:46 am

Let $b_n=a_n-2012$ notice that $a_n$ depends totally on $a_{n-1}$.$a_{n-1}-2012=a_n(a_n-2012)$
So $b_{n+1}=b_n(b_n+2012)$
Then the general formula can be established using induction.I see no better way right now. :mrgreen:

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asif e elahi
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Re: AN INTERESTING PROBLEM BY SAKAL DA

Unread post by asif e elahi » Sat Dec 21, 2013 9:26 pm

zadid xcalibured wrote:Let $b_n=a_n-2012$ notice that $a_n$ depends totally on $a_{n-1}$.$a_{n-1}-2012=a_n(a_n-2012)$
So $b_{n+1}=b_n(b_n+2012)$
Then the general formula can be established using induction.I see no better way right now. :mrgreen:
How to solve thus recurrent relations?

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