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x_1000 pentru un sir recurent

 
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Radu Titiu
Thales


Joined: 28 Sep 2007
Posts: 181
Location: Mures \Bucuresti

PostPosted: Sun Jun 26, 2011 10:32 pm    Post subject: x_1000 pentru un sir recurent Reply with quote

Fie sirul definit prin x_0=5 si recurenta x_{n+1}=x_n+\frac{1}{x_n} .Aratati ca 45 \leq x_{1000} \leq 45,1.
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rgologan



Joined: 19 Jan 2016
Posts: 2

PostPosted: Tue Jan 19, 2016 12:26 pm    Post subject: Reply with quote

Avem $x_n^2=x_{n-1}^2+\frac1{x_n^2}+2$. Prin sumare
$x_n^2=2n+25+\sum_{k=0}^{n-1}\frac1{x_k}^2>2n+25$, apoi
\^n aceea\3i rela\c tie
\[x_n^2<2n+25+\sum_{k=0}^{n-1}\frac1{(2k+25)^2},\]
de unde rezultatul.[/tex]
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enescu
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Joined: 20 May 2008
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PostPosted: Wed Jan 20, 2016 12:42 am    Post subject: Reply with quote

O problemă foarte frumoasă, propusă de Yugoslavia la OIM 1975.
Prof. Cuculescu o prezintă excelent Ón cartea sa despre olimpiadele internaţionale.

A se vedea https://app.box.com/s/8w96zxdw0040rnnqu3e5hwaifzy547qe
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