Problem 33 - Calculus for Biology and Medicine

Question

Find an expression for \(a_{n}\) on the basis of the values of \(a_{0}, a_{1}, a_{2}, \ldots .\) $$ -\frac{1}{2}, \frac{1}{3},-\frac{1}{4}, \frac{1}{5},-\frac{1}{6}, \ldots $$

Step-by-Step Solution

Answer

The expression is \(a_n = \frac{(-1)^n}{n+2}\).

Step 1: Observe the Pattern

Let's first list out the sequence given: \(-\frac{1}{2}, \frac{1}{3}, -\frac{1}{4}, \frac{1}{5}, -\frac{1}{6}, \ldots \) . We need to observe the pattern from the sequence.

Step 2: Identify the Alternating Signs

Notice that the signs of the terms in the sequence alternate: negative, positive, negative, positive, and so on. This alternation can be expressed with \((-1)^n\), where the sign is negative for even \(n\) and positive for odd \(n\).

Step 3: Recognize the Denominator Pattern

The denominators are consecutive integers starting from 2: 2, 3, 4, 5, 6, etc. If we start counting from \(a_0\), we see that the nth term has 2 added to the index: \((n+2)\).

Step 4: Formulate the General Expression

Combining the alternating sign and the pattern of denominators, the expression for the nth term \(a_n\) can be written as: \(a_n = \frac{(-1)^n}{n+2}\).

Step 5: Verify the Expression

Check the first few terms using the formula \(a_n = \frac{(-1)^n}{n+2}\): - For \(n = 0\), \(a_0 = \frac{(-1)^0}{0+2} = \frac{1}{2}\) (and it's negative), so \(-\frac{1}{2}\).- For \(n = 1\), \(a_1 = \frac{(-1)^1}{1+2} = \frac{-1}{3}\).- For \(n = 2\), \(a_2 = \frac{(-1)^2}{2+2} = \frac{1}{4}\) (and it's negative), so \(-\frac{1}{4}\).The pattern matches the given sequence.

Key Concepts

Understanding Alternating Series

An alternating series is a sequence where the terms change sign alternately. In our example, we have the sequence \(-\frac{1}{2}, \frac{1}{3}, -\frac{1}{4}, \frac{1}{5}, -\frac{1}{6}, \ldots \). Here, every other term is negative, and the terms in between are positive. This alternating pattern is crucial in identifying the nature of the series.

This alternation happens because of the factor \((-1)^n\). This factor determines the sign of each term in the sequence:

• If \(n\) is even, \((-1)^n = 1\) and the term is positive.
• If \(n\) is odd, \((-1)^n = -1\) and the term is negative.

Such a simple yet powerful feature distinguishes alternating series from others. Understanding this gives students a straightforward tool to handle complex problems in sequences and series.

Recognizing Patterns in Sequences

Recognizing patterns is an essential skill when working with sequences. For the given sequence, each term follows not only an alternating sign pattern but also a clear pattern in the denominator. The denominators form a straightforward sequence of consecutive integers starting from 2: \(2, 3, 4, 5, 6, \ldots\).

Let's break down what we observe:

• The first term with an index of 0 corresponds to a denominator of 2.
• As we progress, the term index increases by 1 and the denominator by 1 as well.
• Therefore, the nth term of this sequence has a denominator of \(n+2\).

By identifying these patterns, students can simplify finding the general term of complex sequences by using known formulas and relationships.

Formulating the General Term

To express a sequence using a formula—a general term—is a key concept in sequences. It enables one to easily find any term in the sequence without listing all prior terms.

For the example sequence, our task was to find the expression for \(a_n\). By combining the components we've observed:

• The sign of each term is dictated by \((-1)^n\).
• The denominator of each term progresses as \(n+2\).

We can put these together to form the general expression:

\[ a_n = \frac{(-1)^n}{n+2} \]
This formula neatly captures both the pattern of alternating signs and the sequence of denominators. Using this formula, you can easily find any nth term in the sequence, saving time and efficiently addressing sequence problems in calculus.