Sa se calculeze \( \lim_{n \to \infty}\left{\left(45+\sqrt{2007}\right)^n\right} \) unde \( a\in\mathbb{R} \) si \( \left{a\right} \) este partea fractionara a lui "a".
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Concurs "Teodor Topan" - problema 1
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Radu Titiu
- Thales
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Fie \( a=45+\sqrt{2007} \) si \( b=45-\sqrt{2007} \)
Deoarece \( a^n+b^n \in \mathbb{Z} \Leftrightarrow \{a^n\}+[a^n]+\{b^n\}+[b^n] \in \mathbb{Z} \Rightarrow \{a^n\}+\{b^n\}=1 \)(observand ca \( \{a} \) si \( \{b} \) nu pot fi 0)
Dar tinand cont de faptul ca \( b \in (0,1) \) avem
\( \lim_{n\rightarrow \infty} \{a^n\}= \lim_{n\rightarrow \infty} (1-b^n)=1 \)
Deoarece \( a^n+b^n \in \mathbb{Z} \Leftrightarrow \{a^n\}+[a^n]+\{b^n\}+[b^n] \in \mathbb{Z} \Rightarrow \{a^n\}+\{b^n\}=1 \)(observand ca \( \{a} \) si \( \{b} \) nu pot fi 0)
Dar tinand cont de faptul ca \( b \in (0,1) \) avem
\( \lim_{n\rightarrow \infty} \{a^n\}= \lim_{n\rightarrow \infty} (1-b^n)=1 \)
A mathematician is a machine for turning coffee into theorems.