$ {a_0}=2,$
$ {a_n}=2012+{a_0}{a_1}{a_2}{a_3}...{a_{n-2}}{a_{n-1}} $ .
Find out $ {a_{2012}} $.
AN INTERESTING PROBLEM BY SAKAL DA
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MATHPRITOM
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Re: AN INTERESTING PROBLEM BY SAKAL DA
The answer is:
$a_{2012}=2012+2^{2^{2011}}$
$a_{2012}=2012+2^{2^{2011}}$
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$
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MATHPRITOM
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Re: AN INTERESTING PROBLEM BY SAKAL DA
Hi,little brother ,plz, give the full solution.
Re: AN INTERESTING PROBLEM BY SAKAL DA
upssssss........sorrry..........got the mistake probably 
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$
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zadid xcalibured
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Re: AN INTERESTING PROBLEM BY SAKAL DA
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.
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.
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asif e elahi
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Re: AN INTERESTING PROBLEM BY SAKAL DA
How to solve thus recurrent relations?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.