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

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

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Answer 1 · 2016-01-21T08:42:38

$\frac{8}{3}$ $8/3$

Explanation:

$\frac{a_{2}}{a_{1}} = \frac{- 2}{4} = - \frac{1}{2}$ $a_2/a_1=(-2)/4=-1/2$

$\frac{a_{3}}{a_{2}} = \frac{1}{- 2} = - \frac{1}{12}$ $a_3/a_2=1/-2=-1/12$

$\Rightarrow$ $implies$ common ratio $= r = - \frac{1}{2}$ $=r=-1/2$ and first term $= a_{1} = 4$ $=a_1=4$

Sum of infinite geometric series is given by

$S u m = \frac{a_{1}}{1 - r}$ $Sum=a_1/(1-r)$

$\Rightarrow S u m = \frac{4}{1 - (- \frac{1}{2})} = \frac{4}{1 + \frac{1}{2}} = \frac{8}{2} + 1 = \frac{8}{3}$ $implies Sum=4/(1-(-1/2))=4/(1+1/2)=8/2+1=8/3$

$\Rightarrow S = \frac{8}{3}$ $implies S=8/3$

Hence the sum of the given given geometric series is $\frac{8}{3}$ $8/3$ .

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

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