How do you find the sum of the infinite geometric series 1/4 + 1/8 + 1/16 + 1/32 + ..?

Question

How do you find the sum of the infinite geometric series 1/4 + 1/8 + 1/16 + 1/32 + ..?

1 Answer

Impact of this question

17313 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-12T06:20:16

$\frac{1}{2}$ $1/2$

Explanation:

In a geometric series, we multiply by some number $r$ $r$ to get to the next term.

The trick is to find this number $r$ $r$ . If $| r | < 1$ $|r| < 1$ then you can use the following expression to find the sum:
$\frac{a}{1 - r}$ $a/(1-r)$ , where $a$ $a$ is the first term of the series.

we know this:
$\frac{1}{4} \cdot r = \frac{1}{8}$ $1/4 * r = 1/8$
$r = \frac{1}{8} \cdot 4$ $r = 1/8 * 4$
$r = \frac{1}{2}$ $r = 1/2$

Since $| r | < 1$ $|r| < 1$ , we may continue...
$a = \frac{1}{4}$ $a=1/4$

$\frac{a}{1 - r} = \frac{\frac{1}{4}}{1 - \frac{1}{2}}$ $a/(1-r)=(1/4)/(1-1/2)$
$\frac{a}{1 - r} = \frac{\frac{1}{4}}{\frac{1}{2}}$ $a/(1-r)=(1/4)/(1/2)$
$\frac{a}{1 - r} = (\frac{1}{4}) \div (\frac{1}{2})$ $a/(1-r)=(1/4)-:(1/2)$
$\frac{a}{1 - r} = (\frac{1}{4}) \cdot (2)$ $a/(1-r)=(1/4)*(2)$

$\frac{a}{1 - r} = \frac{1}{2}$ $a/(1-r)=1/2$

Science

Math

Humanities

... and beyond

How do you find the sum of the infinite geometric series 1/4 + 1/8 + 1/16 + 1/32 + ..?

1 Answer

Explanation:

Related questions

Impact of this question