Series/Sequence Problem

**Imuell1** · December 9th 2009, 12:49 PM

I'm having a problem with a question

The question is: A chemical plant discharges D units of chemical molecules into a river each week. At the end of the week 90% of the chemical molecules have dissipated.

a) Find the number of units of the molecule after n weeks: I got this
an=D+.01(an-1) or an=a1+.01(an-1)

b) Estimate the amount of the molecule in the river for a very long time.
I know it involves the series of what I found in part a but how do I write that.

c) If the toxic level of the molecule is T units, how large an amount of the molecule can the plant discharge each week?
I don't know what do here.

Any ideas will help, Thanks

**TheEmptySet** · December 9th 2009, 04:50 PM

Originally Posted by Imuell1

I'm having a problem with a question

The question is: A chemical plant discharges D units of chemical molecules into a river each week. At the end of the week 90% of the chemical molecules have dissipated.

a) Find the number of units of the molecule after n weeks: I got this
an=D+.01(an-1) or an=a1+.01(an-1)

b) Estimate the amount of the molecule in the river for a very long time.
I know it involves the series of what I found in part a but how do I write that.

c) If the toxic level of the molecule is T units, how large an amount of the molecule can the plant discharge each week?
I don't know what do here.

Any ideas will help, Thanks

$A_1=D$
$A_2=D+\frac{1}{10}\left(A_1 \right) = D+\frac{1}{10}\left(D\right)= D\left(1+\frac{1}{10}\right)$
$A_3=D+\frac{1}{10}\left(A_2 \right) = D\left(1+\frac{1}{10}+\frac{1}{100}\right)$
If we continue the pattern we get
$A_n=D\sum_{i=1}^{n}\frac{1}{10^{n-1}}$

This should get you started good luck.

**Imuell1** · December 9th 2009, 05:39 PM

Originally Posted by TheEmptySet

$A_1=D$
$A_2=D+\frac{1}{10}\left(A_1 \right) = D+\frac{1}{10}\left(D\right)= D\left(1+\frac{1}{10}\right)$
$A_3=D+\frac{1}{10}\left(A_2 \right) = D\left(1+\frac{1}{10}+\frac{1}{100}\right)$
If we continue the pattern we get
$A_n=D\sum_{i=1}^{n}\frac{1}{10^{n-1}}$

This should get you started good luck.

Thanks.

But for part c...

Would I test for convergence with the p test or something?
or do I somehow take the limit then just set that to less than T?

**Prove It** · December 9th 2009, 05:47 PM

Originally Posted by TheEmptySet

$A_1=D$
$A_2=D+\frac{1}{10}\left(A_1 \right) = D+\frac{1}{10}\left(D\right)= D\left(1+\frac{1}{10}\right)$
$A_3=D+\frac{1}{10}\left(A_2 \right) = D\left(1+\frac{1}{10}+\frac{1}{100}\right)$
If we continue the pattern we get
$A_n=D\sum_{i=1}^{n}\frac{1}{10^{n-1}}$

This should get you started good luck.

I believe it would actually be

$A_n = D\sum_{i = 1}^n\frac{1}{10^{i - 1}}$ .

This is a geometric series, with $a = 1, r = \frac{1}{10}$ and $n = n$ .

So $A_n = \frac{Da(1 - r^n)}{1 - r}$ .

**Imuell1** · December 10th 2009, 07:12 AM

Originally Posted by Prove It

I believe it would actually be

$A_n = D\sum_{i = 1}^n\frac{1}{10^{i - 1}}$ .

This is a geometric series, with $a = 1, r = \frac{1}{10}$ and $n = n$ .

So $A_n = \frac{Da(1 - r^n)}{1 - r}$ .

So I would take the limit as n approaches infinity of $\frac{Da(1 - r^n)}{1 - r}$ ? Which I got 1.11D.

Then for part C would I just set that limit to less than T?

I think I've got it, since the limit = 1.11D, you set that to T and solve for D? giving D=9T/10

**RockHard** · December 10th 2009, 03:28 PM

Note the geometric converges for common ratio's less than 1, its the one of its most known properties. your common ration here is 1/10

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