Geometric Series Question

you are geometrically diluting/mixing 0.1 g of powder A with 100g of powder B, how many times do you have to mix the 2 together to finish the process?

*each time you can only mix an equal portion of powder B to what you currently have mixed.

Eg.

1:1

2:2

4:4

**3. The attempt at a solution**

So the first time you mix them it will be

0.1 g + 0.1 g --> 2^0

then

0.2 g + 0.2 g --> 2^1

0.4 g + 0.4 g -->2^2???

so with the formula is it? 101.1 = 0.1*2^n ???

I don't know what to do from here?

help??

Re: Geometric Series Question

Quote:

Originally Posted by

**mikemd** , how many times do you have to mix the 2 together to finish the process?

How do you know when you are finished?

Quote:

Originally Posted by

**mikemd** , how many times do you have to mix the 2 together to finish the process?

*each time you can only mix an equal portion of powder B to what you currently have mixed.

So after the first mix you have 100.1g in total, then you can add another 100.1g of powder being giving 200.2g in total which 0.1g will be powder A?

Re: Geometric Series Question

Kind of an oddly-worded question. I *think* you almost have it.

Maybe,

0.1*(2+4+8+16+...+What) >= 100.1

0.1*(2^1 + 2^2 + 2^3 + 2^4 +...+ 2^n) >= 100.1

(2^1 - 2^(n+1))/(1-2) >= 1001

Re: Geometric Series Question

Quote:

Originally Posted by

**pickslides** How do you know when you are finished?

Basically you start with 0.1 g and you mix it with 0.1 g of powder B -------> 0.2 g total

take 0.2 g of powder B and mix it with your 0.2 g -----> 0.4 g total

take 0.4 g of powder B and mix it with you 0.4 g -----> 0.8 g

continue untill you used up all of the 100 g of powder B, so I have to figure out how many "mixes" I did before getting to the final mass of 100 g powder B + 0.1 g powder A.

Thanks

Re: Geometric Series Question

Quote:

Originally Posted by

**TKHunny** Kind of an oddly-worded question. I *think* you almost have it.

Maybe,

0.1*(2+4+8+16+...+What) >= 100.1

0.1*(2^1 + 2^2 + 2^3 + 2^4 +...+ 2^n) >= 100.1

(2^1 - 2^(n+1))/(1-2) >= 1001

Thanks TKHunny, which formula did you used (or do you have a link?)?

thanks

Re: Geometric Series Question

I try not to use formulas, *per se*, for this sort of thing. You must need to know how to add a finite geometric series. That's why I left it in that ugly form, so you could see it.

Re: Geometric Series Question

Re: Geometric Series Question

Well, most truthfully, I didn't get anything. Having established the reliable procedure, there was nothing left for me to do. It's your homework.

Let's see: $\displaystyle \frac{2 - 2^{n+1}}{1-2} = 2\cdot \left(2^{n}-1\right) = 1001$ and n = 8.97.

This suggests you may have missed somehting - perhaps that the original exponent was "n+1" rather than "n". Fix that.

In any case, that is not all. We must then answer the question.

"how many times do you have to mix the 2 together to finish the process?"

Ask yourself, "Can I REALLY mix things 0.97 times?" I'm thinking you need only whole numbers. What do you think?

Re: Geometric Series Question

I tried a different approach using

*an* = *ar^*(*n* – 1)

so treating it as

0.2, 0.4, 0.8, 1.6.....100.1

plugging in a = 0.2, an = 100.1 and r = 2. I ended with with the 9.9 value. Would this approach work as well?