Problem 3 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Find the generic terms for the following sums, and write the sums using \(\Sigma\) notation. a. \(2^{2}+4^{2}+6^{2}+8^{2}+10^{2}\) b. \(\quad \frac{1}{1.1}+\frac{1}{1.2}+\frac{1}{1.3}+\frac{1}{1.4}+\frac{1}{1.5}+\frac{1}{1.6}+\frac{1}{1.7}+\frac{1}{1.8}+\frac{1}{1.9}+\frac{1}{2}\) c. \(\left[\left(\frac{1}{1}\right)^{2}+\left(\frac{1}{1.1}\right)^{2}+\left(\frac{1}{1.2}\right)^{2}+\cdots+\left(\frac{1}{1.8}\right)^{2}+\left(\frac{1}{1.9}\right)^{2}\right] \times \frac{1}{10}\) d. \(\quad\left[(-1.0)^{3}+(-0.9)^{3}+(-0.8)^{3}++\cdots+(-0.2)^{3}+(-0.1)^{3}\right] \times \frac{1}{10}\) e. \(\left[\sqrt{1-0.1^{2}}+\sqrt{1-0.2^{2}}+\sqrt{1-0.3^{2}}+\cdots+\sqrt{1-0.9^{2}}+\sqrt{1-1.0^{2}}\right] \times \frac{1}{10}\) f. \([\sqrt{0.1}+\sqrt{0.2}+\sqrt{0.3}+\cdots+\sqrt{1.9}+\sqrt{2.0}] \times \frac{1}{10}\)

Step-by-Step Solution

Answer

a. \(\sum_{n=1}^{5} 4n^2\); b. \(\sum_{n=1}^{10} \frac{1}{1+0.1n}\); c. \(\frac{1}{10} \sum_{n=0}^{9} \frac{1}{(1+0.1n)^2}\); d. \(\frac{1}{10} \sum_{n=0}^{9} (-(1-0.1n))^3\); e. \(\frac{1}{10} \sum_{n=1}^{10} \sqrt{1-(0.1n)^2}\); f. \(\frac{1}{10} \sum_{n=1}^{20} \sqrt{0.1n}\)."

Step 1: Identifying the Pattern for Part a

The given sum is \(2^2 + 4^2 + 6^2 + 8^2 + 10^2\). Notice that each number being squared is an even number, specifically \(2, 4, 6, 8, 10\). Each of these numbers can be expressed as \(2n\) for \(n = 1, 2, 3, 4, 5\). The generic term for the sum is \((2n)^2 = 4n^2\). The sum can be expressed in \(\Sigma\) notation as \(\sum_{n=1}^{5} 4n^2\).

Step 2: Identifying the Pattern for Part b

The given sum is \( \frac{1}{1.1} + \frac{1}{1.2} + \cdots + \frac{1}{2} \). Each denominator is of the form \(1 + 0.1n\) starting from \(n = 1\) to \(n = 10\). Thus, the general term is \(\frac{1}{1+0.1n}\) and the sum is expressed as \(\sum_{n=1}^{10} \frac{1}{1+0.1n}\).

Step 3: Identifying the Pattern for Part c

This part involves a squared reciprocal sum from \(\frac{1}{1^2}\) to \(\frac{1}{1.9^2}\), scaled by 10. The sequence is \(\frac{1}{(1+0.1n)^2}\) where \(n\) goes from \(0\) to \(9\). The sum is \(\sum_{n=0}^{9} \frac{1}{(1+0.1n)^2}\) and it is multiplied by \(\frac{1}{10}\), resulting in \(\frac{1}{10} \sum_{n=0}^{9} \frac{1}{(1+0.1n)^2}\).

Step 4: Identifying the Pattern for Part d

This sum deals with cubes of negative decimals from \((-1)^3\) to \((-0.1)^3\). Each term is \((-0.1n)^3\) for \(n = 10, 9, \, 8, \, \ldots, 1\). The sequence can be expressed and reversed as \((-(1-0.1n))^3\) starting from \(n = 0\) to \(9\). The sum's notation is \(\frac{1}{10}\sum_{n=0}^{9} (-(1-0.1n))^3\).

Step 5: Identifying the Pattern for Part e

In part e, each term takes the form \(\sqrt{1-n^2}\) starting from \(n=0.1\) to \(n=1.0\) in steps of \(0.1\). Hence, the general term is \(\sqrt{1-(0.1n)^2}\) for \(n = 1\) to \(10\). Thus, the sum in \(\Sigma\) notation is \(\frac{1}{10} \sum_{n=1}^{10} \sqrt{1-(0.1n)^2}\).

Step 6: Identifying the Pattern for Part f

For part f, each term is \(\sqrt{n}\) starting from \(n=0.1\) to \(n=2.0\) in increments of \(0.1\). The sequence can be expressed as \(\sqrt{0.1n}\) for \(n=1\) to \(20\). Thus the sum using \(\Sigma\) notation is \(\frac{1}{10} \sum_{n=1}^{20} \sqrt{0.1n}\).

Key Concepts

Generic Terms

When we talk about generic terms in mathematical sequences or series, we are referring to a formula that describes each term individually using a variable, often noted as \( n \). This is important because it allows you to find any specific term in the series without writing out all the preceding terms. For the sum \(2^2 + 4^2 + 6^2 + 8^2 + 10^2\), the generic term of each squared number can be expressed as \((2n)^2 = 4n^2\), where \( n \) ranges from 1 to 5. This means for \( n = 1 \), the term would be \( 4 \times 1^2 = 4 \), and for \( n = 5 \), the term would be \( 4 \times 5^2 = 100 \). Understanding the generic term helps you write linear patterns in compact form.

Pattern Recognition

Pattern recognition is the ability to identify regularities or trends in data. In solving sum problems, recognizing the pattern is the first and crucial step. For example, in the sequence \( \frac{1}{1.1}, \frac{1}{1.2}, \cdots, \frac{1}{2} \), the pattern of the denominators reveals an increment of \(0.1\). Each term can be described generically by \( \frac{1}{1 + 0.1n} \). Grasping the pattern allows you to effectively summarize the series using sigma notation. Look for constant additions, multipliers, or transformations applied uniformly across the terms.

Summation

Summation is the process of adding a sequence of numbers. In mathematics, this is often represented using the uppercase Greek letter \( \Sigma \). Sigma notation lets you express a long series of terms in a concise form. For instance, the sum `\( \sum_{n=1}^{10} \frac{1}{1+0.1n} \)` efficiently signifies the addition of all terms ranging from \( n=1 \) to \( n=10 \). Using summation notation simplifies calculations, enabling easy adjustments in upper and lower limits, and is essential in calculus and higher-level math when dealing with infinite series.

Mathematical Series

A mathematical series is essentially a sum of terms expressed in a sequence. It could go on infinitely or have a defined number of terms. By formulating your sum in sigma notation, you create a series expression such as \( \frac{1}{10} \sum_{n=0}^{9} \frac{1}{(1+0.1n)^2} \), summarizing numerous terms into a compact form. Mathematical series are foundational in calculus, particularly their use in infinite series and power series. Whether you're dealing with a finite series to find sums before a specific point, or extending to infinity, understanding how to manipulate and express these terms plays a crucial role in math problem-solving.