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A doctor has prescribed a medication for a patient. The patient will take 180ml of his medication every 6h and at the same time - every 6h his body will get rid of 60% of the total volume of medication within his body. What's the maximum volume of the medication that patient's body could receive?

See the result below, but I am interested in figuring out how to get that by hand, even if you have a different approach.

I am open for suggestions and any help is greatly appreciated. Thanks.

$$S = \sum_{n=1}^{\infty} 180 \cdot \left(\frac25\right) ^n = 120$$

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    $\begingroup$ Are you asking how to evaluate the sum with hand? $\endgroup$ – Mostafa Ayaz Mar 26 at 20:47
  • $\begingroup$ Note that it doesn't make sense to have $$S_n = \sum_{n=1}^\infty\cdots$$ because $n$ is used as a temporary variable in the sum, whereas $n$ is fixed on the left-hand side. $\endgroup$ – Théophile Mar 26 at 20:48
  • $\begingroup$ @MostafaAyaz: Yes. $\endgroup$ – madjoe Mar 26 at 20:50
  • $\begingroup$ @Théophile: You are absolutely right. I will edit it. $\endgroup$ – madjoe Mar 26 at 20:51
  • $\begingroup$ It’s geometric. $\endgroup$ – let's have a breakdown Mar 26 at 21:31
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Simply use the results for the sum of an infinite geometric series! Recall that

$$1 + x + x^2 + ... = \sum_{k=0}^\infty x^k = \frac{1}{1-x} \;\;\; \text{provided} \; |x| < 1$$

Then notice: you can factor the $180$ out of the sum, as below:

$$S = \sum_{n=1}^\infty 180 \cdot (2/5)^n = 180 \sum_{n=1}^\infty (2/5)^n$$

Be careful now: the infinite geometric sum starts at the zeroth, not the first, term. But we can add $(2/5)^0 = 1$ to the summation to start at $n=0$ and then subtract it. Thus,

$$S = 180 \sum_{n=1}^\infty (2/5)^n = 180 \left(-1 + \sum_{n=0}^\infty (2/5)^n \right) = -180 + \sum_{n=0}^\infty (2/5)^n$$

Then, applying the formula above,

$$S = -180 + 180 \left( \frac{1}{1 - (2/5)} \right) = -180 + 180 \cdot \frac 5 3 = 120$$

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  • $\begingroup$ Thank you, that was also a very clear approach with a full description. Great! $\endgroup$ – madjoe Mar 26 at 21:15
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Well, first the patient takes $180$ mL. Six hours later, $60$ percent, or $108$ mL, has left the patient's system, leaving $72$ mL.

Then the patient takes another $180$ mL, yielding $252$ mL. Six hours later, $60$ percent, or $151.2$ mL, has left the patient's system, leaving $100.8$ mL.

Then the patient takes another $180$ mL, yielding $280.8$ mL. Six hours later, $60$ percent, or $168.48$ mL, has left the patient's system, leaving $112.32$ mL.

If $x_k$ is the amount of medication in the patient's body right before the $k$th dose, then there is $x_k+180$ mL of medication right after the $k$th dose, of which only

$$ x_{k+1} = 0.4(x_k+180) $$

remains after the six hours just before the $k+1$th dose. At equilibrium amount $x$ (to be rigorous, one has to show that there is an equilibrium, but for your purposes, you might not need to do this), we simply equate the two:

$$ x = 0.4(x+180) $$

Solve for $x$ and you get the equilibrium amount just before the next dose. It's unclear to me from your wording of the problem exactly what quantity is desired, but apparently this is it.

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  • $\begingroup$ There is not a desired quantity needed to be defined. This is it, the answer without using the sum. Thank you! $\endgroup$ – madjoe Mar 26 at 21:03
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Hint

$$S=180\cdot\left[({2\over 5})^1+({2\over 5})^2+({2\over 5})^3+\cdots\right]$$also$${5\over 2}S=180\cdot\left[({2\over 5})^0+({2\over 5})^1+({2\over 5})^2+\cdots\right]=180+S$$

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  • $\begingroup$ Looks very simple and easy to understand, brilliant. Thank you! $\endgroup$ – madjoe Mar 26 at 20:58
  • $\begingroup$ Your welcome. Good luck!! $\endgroup$ – Mostafa Ayaz Mar 26 at 20:59
  • $\begingroup$ Hmmm let's try this then $S=1+2+4+...$ and $\frac12 S=\frac12 + 1 + 2 + ...=\frac12 + S \implies S=-1$. This is valid right? $\endgroup$ – Peter Foreman Mar 26 at 21:03
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    $\begingroup$ Nope! Since $|2|>1$ applying this rule is not possible in that way. Furthermore $S=-1$ isn't the only answer. Also $S=\infty$ is another one since$${1\over 2}\infty={1\over 2}+\infty$$but why a same argument (an extra answer $\infty$) doesn't hold for $(0.4)^1+(0.4)^2+(0.4)^3+\cdots$? You might wanna think about it... $\endgroup$ – Mostafa Ayaz Mar 26 at 21:08

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