Amount of Pharmaceutical Left in the Patient's Body After the nth Dose is Administered A patient is administered 500 mg of a certain pharmaceutical every 6 hours. The half-life of the pharmaceutical in the body is 130 minutes. Determine the amount P(n) of the pharmaceutical that remains immediately after the patient takes the nth dose.
Using 
$$
P = P_{0}e^{-kn} \\
$$
With
$$
P_0 = 500 \\
k = \frac{\ln 2 }{ T_{1/2}}\\
k = \frac{\ln 2}{114}
$$
I get
$$
P(n) = 500e^{-0.0053319013889*n}
$$
But it is wrong...
 A: An amount $A$ decays in $6$ hours to $cA$. Your solution shows that you know how to compute $c$.
Immediately after the first dose, the amount is $500$. 
Immediately after the second dose, the amount is $500+500c$, since we have $500$ fresh stuff and $500c$ remaining from the first dose. 
Immediately after the third dose, we have $500+(500+500c)c$, which is $500+500c+500c^2$.
We have $500$ fresh stuff, and the amount immediately after the second dose has been multiplied by $c$.
For immediately after the fourth dose, we take $500$ and add $c$ times the previous answer.
Continue. Immediately after the $n$-th dose we have 
$$500+500c+500c^2+\cdots +500c^{n-1}.$$
Using the formula for the sum of a finite geometric progression, we can write the amount as 
$$\dfrac{500(1-c^n)}{1-c}.$$
Remark: Your calculation followed only what remained of the first dose, and did not take into account the $500$ mg administered every $6$ hours.  Note that if the person continues to take the drug forever, the amount immediately after a dose approaches the value $\frac{500}{1-c}$.
