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 \( a_1^{2002}>a_1\cdot{a_2}\cdot{a_3}\cdot{a_4}\cdot...\cdot{a_{2001}}\cdot{a_{2002}} \).
Giurgiu 2002
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Andi Brojbeanu
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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)\( a_1^{2002}=[(\frac{1}{2})^3]^{2002}=(\frac{1}{2})^{6006}. \)
\( a_1\cdot a_2\cdot a_3\cdot ......\cdot a_{2002} \)\( =(\frac{1}{2})^3\cdot (\frac{1}{2})^4\cdot (\frac{1}{2})^5\cdot .....\cdot (\frac{1}{2})^{2004}=(\frac{1}{2})^{3+4+5+...+2004}=(\frac{1}{2})^{\frac{2004\cdot 2005}{2}-3}=(\frac{1}{2})^{2009007} \).
\( a_1^{2002}>a_1\cdot a_2\cdot a_3\cdot ......\cdot a_{2002}\Leftrightarrow (\frac{1}{2})^{6006}>(\frac{1}{2})^{2009007}\Leftrightarrow 2^{6006}<2^{2009007}, (A) \).
\( a_2=\frac{1}{2}(\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3})=\frac{1}{2}\cdot a_1 \); \( a_3=\frac{1}{2}({\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}})=\frac{1}{2}\cdot a_2 \);.....\( a_n=\frac{1}{2}(\frac{1}{2^{n-1}}-\frac{1}{2^n}-\frac{1}{2^{n+1}})=\frac{1}{2}\cdot a_{n-1} \).
\( a_1=\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}=\frac{1}{2}(1-\frac{1}{2}-\frac{1}{2^2}) \)\( =\frac{1}{2}\cdot \frac{1}{2}{2-1-\frac{1}{2}}=\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=(\frac{1}{2})^3\Rightarrow a_n=(\frac{1}{2})^{n+2}=\frac{1}{2^{n+2}} \)
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)\( a_1^{2002}=[(\frac{1}{2})^3]^{2002}=(\frac{1}{2})^{6006}. \)
\( a_1\cdot a_2\cdot a_3\cdot ......\cdot a_{2002} \)\( =(\frac{1}{2})^3\cdot (\frac{1}{2})^4\cdot (\frac{1}{2})^5\cdot .....\cdot (\frac{1}{2})^{2004}=(\frac{1}{2})^{3+4+5+...+2004}=(\frac{1}{2})^{\frac{2004\cdot 2005}{2}-3}=(\frac{1}{2})^{2009007} \).
\( a_1^{2002}>a_1\cdot a_2\cdot a_3\cdot ......\cdot a_{2002}\Leftrightarrow (\frac{1}{2})^{6006}>(\frac{1}{2})^{2009007}\Leftrightarrow 2^{6006}<2^{2009007}, (A) \).