Question

express the repeating decimal 0.23232323... as a geometric series, and write its sum as the ratio of two integers.

Answer 1

\[.232323...=23\times (\frac{1}{100}+\frac{1}{100^2}+\frac{1}{100^3}+...\] is one way to write is as a geometric series

Answer 2

not to hard to solve for the fraction, even without using a geometric series
\[x=.232323...\\
100x=23.232323...\\99x=23\\x=\frac{23}{99}\]

Answer 3

or you can use
\[23\times \frac{\frac{1}{100}}{1-\frac{1}{100}}\]
\[=23\times \frac{\frac{1}{100}}{\frac{99}{100}}\]
\[=23\times \frac{1}{99}\]
\[=\frac{23}{99}\] either way

Answer 4

or you can remember that if it is a repeating decimal where two digits repeat, it is that number over 99

Answer 5

always?

Answer 6

if it is something like \[.121212...\] then yes, you get
\[\frac{12}{99}=\frac{4}{33}\] you can check it with a calculator

Answer 7

if 3 digits repeat, divide by 999 like
\[.123123123...=\frac{123}{999}\]

Answer 8

if one digit repeats, divide by 9 for example \[.3333...=\frac{3}{9}=\frac{1}{3}\]

Answer 9

why 99??

Answer 10

the short answer is because we write out decimals in base ten
the long answer is that is what you get when you sum the geometric series
see the calculation above and see that you get \(\frac{1}{99}\)

Answer 11

if that is not a satisfactory answer, we can do it slowly and see why it is true

Answer 12

okay so, that is how you write down the sum of the ratios?
each power increases by 1?

Answer 13

\[.23=\frac{23}{100}\\.0023=\frac{23}{10000}=\frac{23}{100^2}\\
.000023=\frac{23}{100^3}\] etc

Answer 14

the font is so weird, your 3 looks like a two haha. okay i see now!

Answer 15

factor out the \(23\) and the remaining sum is
\[\frac{1}{100}+\frac{1}{100^2}+\frac{1}{100^3}+...\] if you are familiar with sigma notation, it is the same as
\[\sum_{k=1}^{\infty}(\frac{1}{100})^k\]

Answer 16

yeah it is large on my screen but i have heard that before

Answer 17

oooh okay i see.

Answer 18

oh haha yah it looks normal now that i made my screen bigger lol.

Answer 19

would i write the 23 outside of the (1/100)?