# Change repeating decimal to fraction

• Dec 5th 2007, 07:15 PM
oceanmd
Change repeating decimal to fraction
0.143 (3 repeating) I solved the following way:
x=0.143 (3 repeating)
100x=14.3 (3 repeating)
1000x= 143.3 (3 repeting)
1000x - 100x = 143.3-14.3
900x = 129
x= 129/900

My question: how to solve this problem using geometric series?
14/100 + 0.003 (3 repeating)

I guess we need to find fraction for 0.003 (3 reapeting)

What is the first term? What is the common ratio?

Thanks
• Dec 5th 2007, 07:35 PM
Henderson
You've got the $\displaystyle \frac{14}{100}$ just fine.

Then you're adding in .003 + .0003 + .00003 + .000003 + ...,

so your first term is .003, and your common ratio is $\displaystyle \frac{1}{10}$.

Also, it's possible to say that .1433333333... is .333333333... -.2+.01, or $\displaystyle \frac{1}{3} - \frac{1}{5} + \frac{1}{100}$.
• Dec 5th 2007, 07:42 PM
oceanmd
Thank you, got it
• Dec 5th 2007, 07:47 PM
oceanmd
Simplify
Henderson, I see you are online now. Can you please tell me whether 8 taken 7 at a time is 56 (it's on binomial theorem) It was not explained in class, but is included in the test review, I do not know what it is about, just used the formula.

Thank you
• Dec 5th 2007, 07:56 PM
Henderson
Sure-

8 taken 7 at a time is the same as the number of ways to leave one guy out, so there are only going to be 8 ways to do this.

Here's a decent explanation of what you're talking about: Binomial Theorem.

Hope this helps!
• Dec 5th 2007, 08:11 PM
oceanmd
Use Pascal's Triangle and expand
(2x-3)^4 = 16x^4-96x^3+216x^2-216x + 81

Is this correct? I did it using binomial theorem, why do I need to use Pascal's triangle?

Thank you
• Dec 5th 2007, 08:34 PM
Henderson
Looks correct to me.

You know the numbers you're getting from combinations that you're multiplying into each term (in your example, the 1,4,6,4, and 1)? Rather than run a combination for each term, Pascal's triangle gives you the whole list of coefficients (since you raised to the fourth power, these numbers are the fourth row of Pascal)