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'Weekly Challenge 32: Medicine Half Life' printed from http://nrich.maths.org/

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As the question indicates a long-term level I might imagine having taken the tablets for several days and will assume that the tablets are taken in 12-hourly intervals. When I take a tablet, the amount remaining of the previous tablet will have halved, the tablet before that quartered an so on.
 
If I think carefully and clearly about this then I can write down an expression for the total mass $M(n)$ remaining in the system upon taking the $n$th tablet:
 
$$
M(n) = 400\left(1 +\frac{1}{2}+\frac{1}{4}+\dots +\frac{1}{2^{n-1}}\right)mg
$$
(With such an expression I should check that the limits are correct: My formula gives$M(1) = 400(1)$ and $M(2) = 400( 1+\frac{1}{2^1})$, which are correct).
 
 $M(n)$ is a geometric series, which sums to
$$
M(n) = 400\left(2-\frac{1}{2^{n-1}}\right)
$$
 
The long term level is found by taking the large $n$ limit, giving a level of $800$mg in the system, as one might intuitively guess.
 
 
Extension:
If the body breaks down half of the dose every $6$ hours then the levels of the previous tablets will have quartered each time. Thus the relevent geometric series is
$$
M(n) = 400\left(1 +\frac{1}{4}+\frac{1}{16}+\dots +\frac{1}{4^{n-1}}\right)
$$
The sum of this series is
$$
M(n) = 400\left(\frac{1-0.25^n}{1-0.25}\right)
$$
The long term level will be $533\frac{1}{3}$mg