I am giving the problem again with a correction.
Let $a_{0}=1996$ and $a_{n+1}=\frac{a_{n}^2}{a_{n}+1}$ for $n=0,1,2,3,.....$. Prove that, $\boxed{[a_{0}]=1996-n}$ for $n=0,1,2,3,....,999$. Here $[x]$ is the floor of $x$.
Edit: I think the boxed part should be $[a_n]=1996-n$.
Re:determining the floor
Last edited by Masum on Sat Oct 13, 2012 8:42 pm, edited 1 time in total.
Reason: It is better to use a space after the end of line and the \frac command.
Reason: It is better to use a space after the end of line and the \frac command.
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- Phlembac Adib Hasan
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Re: Re:determining the floor
The problem is actually IMO SLT 1994-1 Proposed by USA (NOT Russia).The proof easily follows from induction.
Hint :
Hint :
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