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Giurgiu 2002

 
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alex2008
Leibniz


Joined: 19 Oct 2008
Posts: 489
Location: Tulcea

PostPosted: Sun Nov 09, 2008 9:07 am    Post subject: Giurgiu 2002 Reply with quote

Se considera numerele de forma :
a_1=\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}, a_2=\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}, a_3=\frac{1}{2^3}-\frac{1}{2^4}-\frac{1}{2^5} ...
a) Sa se scrie forma numarului a_{2002} iar apoi sa se calculeze si sa se aseze in ordine crescatoare numerele a_1,a_2,a_3,a_4,...,a_{2002}.
b) Sa se arate ca a_1^{2002}>a_1\cdot{a_2}\cdot{a_3}\cdot{a_4}\cdot...\cdot{a_{2001}}\cdot{a_{2002}}.
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Andi Brojbeanu
Newton


Joined: 22 Mar 2009
Posts: 383
Location: Targoviste (Dambovita)

PostPosted: Mon Apr 20, 2009 8:41 pm    Post subject: Reply with quote

a)a_{2002}=\frac{1}{2^{2002}}-\frac{1}{2^{2003}}-\frac{1}{2^{2004}}
a_2=\frac{1}{2}(\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3})=\frac{1}{2}\cdot a_1; a_3=\frac{1}{2}({\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}})=\frac{1}{2}\cdot a_2;.....a_n=\frac{1}{2}(\frac{1}{2^{n-1}}-\frac{1}{2^n}-\frac{1}{2^{n+1}})=\frac{1}{2}\cdot a_{n-1}.
a_1=\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}=\frac{1}{2}(1-\frac{1}{2}-\frac{1}{2^2})=\frac{1}{2}\cdot \frac{1}{2}{2-1-\frac{1}{2}}=\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=(\frac{1}{2})^3\Rightarrow a_n=(\frac{1}{2})^{n+2}=\frac{1}{2^{n+2}}
Atunci a_2=\frac{1}{2^4} ;a_3=\frac{1}{2^5};.... ;a_{2002}=\frac{1}{2^{2004}.
Al doilea termen este jumatate din primul, al treilea jumatate din al doilea, ....., al 2002-lea termen este jumatate din al 2001-lea.
Deci, ordinea este a_{2002}, a_{2001}, ....., a_3, a_2, a_1.
b)a_1^{2002}=[(\frac{1}{2})^3]^{2002}=(\frac{1}{2})^{6006}.
a_1\cdot a_2\cdot a_3\cdot ......\cdot a_{2002}=(\frac{1}{2})^3\cdot (\frac{1}{2})^4\cdot (\frac{1}{2})^5\cdot .....\cdot (\frac{1}{2})^{2004}=(\frac{1}{2})^{3+4+5+...+2004}=(\frac{1}{2})^{\frac{2004\cdot 2005}{2}-3}=(\frac{1}{2})^{2009007}.
a_1^{2002}>a_1\cdot a_2\cdot a_3\cdot ......\cdot a_{2002}\Leftrightarrow (\frac{1}{2})^{6006}>(\frac{1}{2})^{2009007}\Leftrightarrow 2^{6006}<2^{2009007}, (A).
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