Problem 96 - Precalculus: Mathematics for Calculus

Question

A certain drug is administered once a day. The concentration of the drug in the patient's bloodstream increases rapidly at first, but each successive dose has less effect than the preceding one. The total amount of the drug (in mg) in the bloodstream after the \(n\) th dose is given by $$\sum_{k=1}^{n} 50\left(\frac{1}{2}\right)^{k-1}$$ (a) Find the amount of the drug in the bloodstream after \(n=10\) days. (b) If the drug is taken on a long-term basis, the amount in the bloodstream is approximated by the infinite series \(\sum_{k=1}^{\infty} 50\left(\frac{1}{2}\right)^{k-1} .\) Find the sum of this series.

Step-by-Step Solution

Answer

(a) After 10 days, approximately 99.90234 mg; (b) Infinite series sum is 100 mg.

Step 1: Understanding Geometric Series

The expression \(\sum_{k=1}^{n} 50\left(\frac{1}{2}\right)^{k-1}\) is a geometric series with first term \(a = 50\) and common ratio \(r = \frac{1}{2}\).

Step 2: Calculating Finite Geometric Series for n=10

The formula for the sum \(S_n\) of the first \(n\) terms of a geometric series is given by \(S_n = a \frac{1-r^n}{1-r}\). Substitute \(a = 50\), \(r = \frac{1}{2}\), and \(n = 10\) into the formula: \[S_{10} = 50 \frac{1-\left(\frac{1}{2}\right)^{10}}{1-\frac{1}{2}}\]Calculate \(S_{10}\).

Step 3: Simplifying the Expression for n=10

Start simplifying by computing \(\left(\frac{1}{2}\right)^{10} = \frac{1}{1024}\). Then plug this value back into the sum formula:\[S_{10} = 50 \times \frac{1 - \frac{1}{1024}}{\frac{1}{2}} = 50 \times 2\left(1 - \frac{1}{1024}\right)\]Therefore, \(S_{10} = 100 \times \left(1 - \frac{1}{1024}\right)\).

Step 4: Calculating Infinite Geometric Series

For an infinite series \(\sum_{k=1}^{\infty} ar^{k-1}\), the sum \(S_\infty\) is given by \(S_\infty = \frac{a}{1-r}\) when \(|r| < 1\). Substitute \(a = 50\) and \(r = \frac{1}{2}\):\[S_{\infty} = \frac{50}{1 - \frac{1}{2}} = \frac{50}{\frac{1}{2}} = 100\]This gives us the limiting value as \(n\) approaches infinity.

Step 5: Conclusion and Final Results

From our calculations, we found:- After 10 days, the amount of the drug in the bloodstream is approximately \(100 \times \left(1 - \frac{1}{1024}\right) = \frac{102300}{1024} \) mg.- As \(n\) approaches infinity, the maximum amount possible in the bloodstream stabilizes at 100 mg due to the repeating diminishing returns of each dose.

Key Concepts

Infinite Series

An infinite series is a sum of the terms of an infinite sequence. In mathematics, we frequently analyze infinite series to determine convergence or divergence. Convergence refers to whether the series approaches a specific value as more terms are added.

In the original exercise, the infinite series involved is \(\sum_{k=1}^{\infty} 50\left(\frac{1}{2}\right)^{k-1}\). Calculating this series involves using the concept of geometric series, specifically when the number of terms approaches infinity. Since the common ratio \(r = \frac{1}{2}\) is between \(-1\) and \(1\), the series converges. The formula for the sum of an infinite geometric series \(S_\infty\) is \(\frac{a}{1-r}\).

By plugging in the values, \(a = 50\) and \(r = \frac{1}{2}\), we obtain \(S_\infty = \frac{50}{1-\frac{1}{2}} = 100\). This result represents the sum to which the series converges, illustrating how the impact of each successive dose of the drug decreases over time, eventually stabilizing.

Finite Geometric Series

A finite geometric series involves summing a specific number of terms from a geometric progression. For these series, the sum \(S_n\) can be calculated using the formula \(S_n = a \frac{1-r^n}{1-r}\), where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms.

Regarding our exercise, the finite geometric series for 10 doses is described by the sum \(\sum_{k=1}^{10} 50\left(\frac{1}{2}\right)^{k-1}\). By applying the formula, \(a = 50\), \(r = \frac{1}{2}\), and \(n = 10\), we compute:

• Calculate \(\left(\frac{1}{2}\right)^{10} = \frac{1}{1024}\).
• Use the formula: \(S_{10} = 50 \frac{1 - \frac{1}{1024}}{1 - \frac{1}{2}}\).
• Simplify to find \(S_{10} = 100\left(1 - \frac{1}{1024}\right)\).

Ultimately, summing these terms approximates the drug concentration in the bloodstream after 10 doses, revealing the decreasing effect with each additional dose.

Geometric Progression

A geometric progression (or geometric sequence) is a sequence of numbers where each term is derived by multiplying the previous one by a constant called the common ratio (\(r\)). The formula for terms in a geometric progression is expressed as \(a, ar, ar^2, ar^3, \ldots\), where \(a\) is the first term.

In our exercise, the geometric progression is reflected in the pattern \(50\left(\frac{1}{2}\right)^{0}, 50\left(\frac{1}{2}\right)^{1}, 50\left(\frac{1}{2}\right)^{2}, \ldots\) suggesting how each dose of the drug reduces due to the diminishing response factor of \(\frac{1}{2}\).

• First term: \(a = 50\)
• Common ratio: \(r = \frac{1}{2}\)

This progression forms the basis of the finite and infinite series calculations associated with the drug doses. Understanding this concept helps in grasping how geometric sequences underpin both finite and infinite series, revealing predictable patterns in diminishing returns of drug concentration in the bloodstream over time.