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up:: [[suite géométrique]]
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title:: "$\sum\limits_{k=0}^{n} q^{k} = \dfrac{1-q^{k}}{1-q}$"
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#maths/analyse #maths/arithmétique
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#s/maths/analyse #s/maths/arithmétique
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$\displaystyle\sum\limits_{k= p}^{N} x^{k} = \frac{x^{p} - x^{N+1}}{1 - x} = \frac{(\text{premier terme}) - (\text{premier terme pas pris})}{1-x}$
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