What is the sum of the infinite geometric series 8 + 4 + 2 + 1 +... ?

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What is the sum of the infinite geometric series 8 + 4 + 2 + 1 +... ?

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Answer 1 · 2014-09-11T09:46:17

The common ratio is $\frac{1}{2}$ $1/2$ or $0.5$ $0.5$ . If you multiply the current term by the the common ratio the the output will be the next term.

$8 \cdot \frac{1}{2} = 4$ $8 * 1/2 = 4$

$4 \cdot \frac{1}{2} = 2$ $4*1/2=2$

$2 \cdot \frac{1}{2} = 1$ $2 * 1/2 = 1$

etc ...

because the absolute value of $r$ $r$ is less than 1 we can use the following formula.

$\frac{a}{1 - r}$ $a/(1-r)$ where $a$ $a$ is the first term and $r$ $r$ is the common ratio

In our problem $a = 8$ $a = 8$ and $r = 0.5$ $r=0.5$

Substitute

$\frac{8}{1 - 0.5} = \frac{8}{0.5} = 16$ $8/(1-0.5) = 8/0.5 = 16$

The sum of this infinite geometric series is 16.

Also, another formula you can use that is guaranteed to work every time, no matter what, is:

$S_{n} = a (\frac{r^{n} - 1}{r - 1})$ $S_n=a((r^n-1)/(r-1))$

All the variables work the same way as above, and "n" is the number of terms in the series. So, say you wanted to find the sum of the first 10 terms and were to substitute everything in:

$S_{10} = 8 (\frac{{0.5}^{10} - 1}{0.5 - 1})$ $S_10=8((0.5^10-1)/(0.5-1))$

$S_{10} = 15.984375$ $S_10=15.984375$

Therefore the sum if the series is 15.98!

Hopefully this was helpful! You can use either formula, it's just a matter of preference; the second one is more reliable and accurate though! :)

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What is the sum of the infinite geometric series 8 + 4 + 2 + 1 +... ?

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