Fractii

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Matematica Ajutor :: Cereri si rezolvari de probleme :: Clasa a-VII-a

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Fractii

andy28 Dum Noi 18, 2012 9:50 pm

$\frac{1}{2} - \frac{1}{{{2^2}}} - \frac{1}{{{2^3}}} - \cdots - \frac{1}{{{2^{40}}}}$

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andy28: Mesaje : 86
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$\begin{array}{l} S = \frac{1}{2} - \frac{1}{{{2^2}}} - \frac{1}{{{2^3}}} - \cdots - \frac{1}{{{2^{40}}}} = \underbrace {1 - \frac{1}{2}}_{ = \frac{1}{2}} - \frac{1}{{{2^2}}} - \frac{1}{{{2^3}}} - \cdots - \frac{1}{{{2^{40}}}}\\ = 1 - \frac{1}{2}\left( {1 + \frac{1}{2} + \cdots + \frac{1}{{{2^{39}}}}} \right)\\ {\text{Vom folosi formula: }}1 + q + {q^2} + \cdots + {q^{n - 1}} = \frac{{1 - {q^n}}}{{1 - q}}\\ \left\{ \begin{array}{l} q = \frac{1}{2}\\ n - 1 = 39 \end{array} \right. \Rightarrow S = 1 - \frac{1}{2} \times \frac{{1 - \frac{1}{{{2^{40}}}}}}{{1 - \frac{1}{2}}} = \frac{1}{{{2^{40}}}} \end{array}$