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Just after taking the (n-1)^(th) dose the amount of the drug is a_(n-1). Before the next dose, 60 % of the drug is removed from the bloodstream. Therefore, 40 %=0.4 of the old drug remains. Next, we add 325 milligrams when taking the next dose. ccccc c Amount of the drug after the $n^(th)$ dose & =0.4 * & c Amount of the drug after the $(n-1)^(th)$ dose & + & New dose ⇓ & & ⇓ & & ⇓ a_n & =0.4 * & a_(n-1) & + & 325 Therefore, we can write a recursive rule: a_1=325, a_n=0.4a_(n-1)+325 for n>1.
=Round((0.4)*(A1)+325,2)in cell A2. When we hit enter, we will get the following.
Next, copy cell A2, highlight cells A3 through A30, and paste. Then, note that the numbers stabilize around cell A12.
After cell A11 we will keep getting the same value, 541.66. This tells us that the maintenance level is 541.66.
Recursive Rule:
a_1=650, a_n=0.4a_(n-1)+650forn>1=Round((0.4)*(A1)+650,2)
in cell A2. When we hit enter, we will get the following.
Next, copy cell A2, highlight cells A3 through A30, and paste. Then, note that the numbers stabilize around cell A13.
After cell A12 we will keep getting the same value, 1083.33. This tells us that the maintenance level is 1083.33. The maintenance level of the drug from Part B is 541.66. Since 1083.33541.66≈ 2, the maintenance level doubles when we double the dose.