How do you find the sum of the infinite geometric series 1-1/2+1/4+1/8?

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How do you find the sum of the infinite geometric series 1-1/2+1/4+1/8?

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Answer 1 · 2016-01-25T20:46:51

As Alan P. indicated, this question has to be modified. I'm gonna assume that it is $- \frac{1}{8}$ $-1/8$ instead of $+ \frac{1}{8}$ $+1/8$

Explanation:

First you must find r, the common ratio.

You find this with the following formula: r = $\frac{t_{2}}{t_{1}}$ $t_2/t_1$

r = #(-1/2)/1)

r = $- \frac{1}{2}$ $-1/2$

The ratio is therefore $- \frac{1}{2}$ $-1/2$ . The formula for the sum of an infinite geometric series is $s_{\infty} = \frac{a}{1 - r}$ $s_oo = a/(1 - r)$

$s_{\infty} = \frac{1}{1 - (- \frac{1}{2})}$ $s_oo = 1 / ( 1 - (-1/2)$

$s_{\infty} = \frac{1}{\frac{3}{2}}$ $s_oo = 1/(3/2)$

$s_{\infty} = \frac{2}{3}$ $s_oo = 2/3$

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How do you find the sum of the infinite geometric series 1-1/2+1/4+1/8?

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