Найдите сумму бесконечно убывающей геометрической прогрессии: 3; -1; 1/3;….. . Created by none6. algebra-ru

3;….. ...

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Найдите сумму бесконечно убывающей геометрической прогрессии: 3; -1; 1/3;…..

anel7776

Ответ:

$\{b_{n}\}:\ \ 3\ ;\ -1\ ;\ \dfrac{1}{3}\ ;\ ...\\\\\\q=\dfrac{b_2}{b_1}=\dfrac{-1}{3}=-\dfrac{1}{3}\ \ ,\ \qquad (\ proverim:\ q=\dfrac{b_3}{b_2}=\dfrac{1/3}{-1}=-\dfrac{1}{3}\ )\\\\\\S=\dfrac{b_1}{1-q}=\dfrac{3}{1-(-\frac{1}{3})}=\dfrac{3}{1+\frac{1}{3}}=\dfrac{3}{\frac{4}{3}}=\dfrac{9}{4}=2,25$

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