# Series with repeating decimal

• Oct 13th 2008, 10:28 AM
redman223
Series with repeating decimal
10.9090909091.....

I don't really know what to do with this problem, I know that I should set it up like this:

$
\frac{1090}{100} + (\frac {1090}{100}* \frac {1}{100})+ (\frac {1090}{100}*(\frac{1}{100})^2)....
$

But whats next?

So A=1090/100 and R=1/100?
• Oct 13th 2008, 11:54 AM
mr fantastic
Quote:

Originally Posted by redman223
10.9090909091.....

I don't really know what to do with this problem, I know that I should set it up like this:

$
\frac{1090}{100} + (\frac {1090}{100}* \frac {1}{100})+ (\frac {1090}{100}*(\frac{1}{100})^2)....
$

But whats next?

So A=1090/100 and R=1/100?

Here's a short cut:

S = 10.909090.....

100S = 1090.9090.....

Therefore 100S - S = 1080 => 99S = 1080 => S = 1080/99.

Otherwise, what you do next is sum the infinite geometric series.
• Oct 13th 2008, 12:31 PM
redman223
Will that always work? Do you basically just take the original repeating number s, multiply it by 100, subtract s from it to make it 99s and then divide by 99?

Also, the number was 10.9090909091, the 1 at the end is throwing me off, because I don't know if that trick still applies or not.
• Oct 13th 2008, 12:41 PM
11rdc11
The trick is to count how many repeating decimals you have,

For instance in

$x=0.94444$

Muliply both sides of this equation by $10^n$, where n is the number of digits in the repeating string. Here, the single digit 4 repeats infinitely, so n = 1

$10^1(x)=10^1(.9444444...)$

$10x = 9.4444444...$

Subtract the original equation (x=0.9444444...) from the modified equation (10x = 9.444444...) which equals

9x= 8.5

$x= \frac{17}{18}$
• Oct 13th 2008, 01:45 PM
redman223
Ok, that makes sense. But what about the number in my problem, I am confused by the way its written.

10.9090909091... I don't know what is repeated. Is it the 1? Is is the whole thing? The problem is that there is no teacher or anything to help me clarify that problem.

So I will have to do it multiple times until the computer accepts the answer.

So would I take 10.9090909091 and multiply it by what? By 1? Or by 10?
• Oct 13th 2008, 02:04 PM
Plato
But you may be required to use a series solution.
$\begin{array}{rcl}
{10.90\overline {90} } & = & {10 + .9 + .009 + .00009 \cdots } \\
{} & = & {10 + 9\left( {10^{ - 1} } \right) + 9\left( {10^{ - 3} } \right) + \cdots } \\
{} & = & {10 + 9\sum\limits_{k = 1}^\infty {10^{ - (2k - 1)} } } \\
{} & = & {10 + 9\frac{{10^{ - 1} }}
{{1 - 10^{ - 2} }}} \\ \end{array}
$
• Oct 13th 2008, 02:16 PM
redman223
I am still stuck on what to do, because of the 1 at the end of

10.9090909091
• Oct 13th 2008, 02:53 PM
Plato
Quote:

Originally Posted by redman223
I am still stuck on what to do, because of the 1 at the end of 10.9090909091

I think that is a misprint.
If not the the other methods do not work either.
If it is not a misprint then it is a pointless question.
• Oct 13th 2008, 02:55 PM
WantMath
I think so, too, or they rounded the number by accident
• Oct 13th 2008, 03:12 PM
redman223
Express http://webwork.csufresno.edu/webwork...04dfea90d1.png as a rational number, in the form http://webwork.csufresno.edu/webwork...fce30f9be1.png
where http://webwork.csufresno.edu/webwork...da050b4dd1.png and http://webwork.csufresno.edu/webwork...12a8ea5921.png are positive integers with no common factors.

This is how the problem is written out on webwork (I despise webwork, I wish our school would stop using it). I tried using 1080 and 99 but those didn't work.
• Oct 13th 2008, 04:13 PM
11rdc11
$10 + \frac{90}{100} + \frac{90}{10000} + \frac{90}{1000000} + \frac{90}{100000000} +\frac{91}{10000000000} = \frac{109090909091}{10000000000}$
• Oct 13th 2008, 04:29 PM
redman223
I tried that earlier and it didn't work. I am stumped.
• Oct 13th 2008, 04:32 PM
11rdc11
Quote:

Originally Posted by redman223
I tried that earlier and it didn't work. I am stumped.

Hmm so am i(Wondering)
• Oct 13th 2008, 04:34 PM
mr fantastic
Quote:

Originally Posted by redman223
I tried that earlier and it didn't work. I am stumped.

As Plato remarked earlier, it's probably an error. Can you check whether or not it's an error in the question.

Otherwise it is simple but tedious. And ridiculous (an observation also earlier made).
• Oct 13th 2008, 04:49 PM
mr fantastic
Quote:

Originally Posted by redman223
Express http://webwork.csufresno.edu/webwork...04dfea90d1.png as a rational number, in the form http://webwork.csufresno.edu/webwork...fce30f9be1.png
where http://webwork.csufresno.edu/webwork...da050b4dd1.png and http://webwork.csufresno.edu/webwork...12a8ea5921.png are positive integers with no common factors.

This is how the problem is written out on webwork (I despise webwork, I wish our school would stop using it). I tried using 1080 and 99 but those didn't work.

If it's not a typo in the question here is an outline of the ridiculous solution:

10 + 0.909090909 + 0.0000000001111........ (I assume)

Obviously 0.909090909 = 909090909/1000000000

Let S = 0.00000000011111.....

Then 1000000000S = 0.1111111 and 10000000000S = 1.1111111

Therefore 9000000000S = 1 => S = 1/9000000000.

So your number is 10 + (909090909/1000000000) + (1/9000000000).

Now add the fractions in the usual way to get the required representation.