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

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