Question

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Answer 1

progresie geometrică, cu primul termen 1, rația ½ și cu 2009 termeni

$S_{n} = \dfrac{b_{1}\cdot ({q}^{n} - 1) }{q - 1}$

$S_{2009} = \dfrac{1\cdot \big(\big( \frac{1}{2}\big)^{n} - 1 \big) }{ \frac{1}{2} - 1} = \dfrac{1 - \big(\frac{1}{2}\big)^{n}}{ \frac{1}{2}} = 2\bigg(1 - \dfrac{1}{{2}^{n}}\bigg) = 2 - \dfrac{1}{{2}^{n - 1}}$

$0 < \dfrac{1}{{2}^{n - 1}} < 1 \iff - 1 < - \dfrac{1}{{2}^{n - 1}} < 0$

$1 < 2 - \dfrac{1}{{2}^{n - 1}} < 2$

$\implies S \in (1;2)$