Chapter 8.2, Problem 49E

a.

To determine

To find: what quantity of the drug remains in the body after the patient takes three pills.

Answer to Problem 49E

Quantity of the drug remains in the body after the patient takes three pills is 105.25.

Explanation of Solution

Given information: A patient is prescribed a drug and is told to take one 100 mg pill

every eight hours. After eight hours, about 5% of the drug remains in the body.

Calculation:

The series in this situation is:

  ${\displaystyle \sum _{n=1}^{\infty }100(\frac{1}{20}}{)}^{n-1}$

Sum when n =3.

  ${\displaystyle \sum _{n=1}^{3}100(\frac{1}{20}}{)}^{n-1}=100.{(\frac{1}{20})}^{1-1}+100.{(\frac{1}{20})}^{2-1}+100.{(\frac{1}{20})}^{3-1}=100.(1)+\frac{100}{20}+\frac{100}{400}=105.25$

b.

To determine

To find: what quantity remains after n pills are taken.

Answer to Problem 49E

Quantity remains after n pills are taken:

  $100{\displaystyle \sum _{n=0}^n(\frac{1}{20}}{)}^n$ .

Explanation of Solution

Given information: A patient is prescribed a drug and is told to take one 100 mg pill

every eight hours. After eight hours, about 5% of the drug remains in the body.

Calculation:

Quantity remains after n pills are taken:

  ${\displaystyle \sum _{n=0}^{n}100(\frac{1}{20}}{)}^n=100{\displaystyle \sum _{n=0}^n(\frac{1}{20}}{)}^n$

Where n is the number of pills taken.

c.

To determine

To find: what happens in the long run.

Answer to Problem 49E

105.263 mg.

Explanation of Solution

Given information: A patient is prescribed a drug and is told to take one 100 mg pill

every eight hours. After eight hours, about 5% of the drug remains in the body.

Calculation:

Quantity remains after n pills are taken:

  $100{\displaystyle \sum _{n=0}^n(\frac{1}{20}}{)}^n$ .

Since series is geometric sequence : with $r=\frac{1}{20}$ .

So,

  $100{\displaystyle \sum _{n=0}^n(\frac{1}{20}}{)}^n=100(\frac{1}{1-\frac{1}{20}})=100(\frac{20}{19})=105.263\text{mg}$