Problem 20 - Algebra for College Students

Question

Write the first four terms of the infinite sequence whose nth term is given. \(b_{n}=\frac{1}{(n+1)(n+2)}\)

Step-by-Step Solution

Answer

The first four terms are \( \frac{1}{6}, \frac{1}{12}, \frac{1}{20}, \frac{1}{30} \).

Step 1: Understand the Given Formula

The given formula for the nth term of the sequence is \( b_n = \frac{1}{(n+1)(n+2)} \). To find the first four terms, evaluate this formula for \( n = 1, 2, 3, \) and \( 4 \).

Step 2: Compute the First Term

For \( n = 1 \): Substitute \( n = 1 \) into the formula: \[ b_1 = \frac{1}{(1+1)(1+2)} = \frac{1}{2 \cdot 3} = \frac{1}{6} \]

Step 3: Compute the Second Term

For \( n = 2 \): Substitute \( n = 2 \) into the formula: \[ b_2 = \frac{1}{(2+1)(2+2)} = \frac{1}{3 \cdot 4} = \frac{1}{12} \]

Step 4: Compute the Third Term

For \( n = 3 \): Substitute \( n = 3 \) into the formula: \[ b_3 = \frac{1}{(3+1)(3+2)} = \frac{1}{4 \cdot 5} = \frac{1}{20} \]

Step 5: Compute the Fourth Term

For \( n = 4 \): Substitute \( n = 4 \) into the formula: \[ b_4 = \frac{1}{(4+1)(4+2)} = \frac{1}{5 \cdot 6} = \frac{1}{30} \]

Key Concepts

infinite sequence

An infinite sequence is a list of numbers where a first term is followed by an endless continuation of other terms. Each term in the sequence follows a specific pattern or rule. For example, in the given problem, the sequence is defined by the formula for its nth term as \(b_{n} = \frac{1}{(n+1)(n+2)}\). This means the sequence continues indefinitely, generating new terms as you substitute increasing values of \(n\). It is important to note that the characteristic that defines an infinite sequence is its lack of an ending point. As you progress, each successive term builds upon the previous terms following the rule given by the nth term formula.

nth term formula

The nth term formula is a mathematical expression that allows you to find any term in the sequence. In our exercise, the nth term formula is given by \(b_{n} = \frac{1}{(n+1)(n+2)}\). This formula directly reveals the pattern used to find each term. To get the first few terms, substitute \(n = 1, 2, 3, 4\) into the formula and solve for \(b_n\). For instance:
1. **First Term**: For \(n=1\), \(b_1 = \frac{1}{(1+1)(1+2)} = \frac{1}{6}\)
2. **Second Term**: For \(n=2\), \(b_2 = \frac{1}{(2+1)(2+2)} = \frac{1}{12}\)
3. **Third Term**: For \(n=3\), \(b_3 = \frac{1}{(3+1)(3+2)} = \frac{1}{20}\)
4. **Fourth Term**: For \(n=4\), \(b_4 = \frac{1}{(4+1)(4+2)} = \frac{1}{30}\)
This systematic substitution helps in understanding how each term is formed and the general pattern of the sequence.

sequence evaluation

Sequence evaluation refers to the process of finding certain terms of the sequence based on a given formula. It typically involves substituting different values for \(n\) in the nth term formula and calculating the results. Evaluation steps clarify how each term materializes from the formula. For example, in the given sequence, if we evaluate for \(n = 1\), the term is \(b_1 = \frac{1}{6}\). Continuing with \(n = 2\), we get \(b_2 = \frac{1}{12}\). Each calculation follows the substitution and simplification steps provided by the formula. Through sequence evaluation, you can gain insights into the characteristics and behavior of the sequence as it progresses, thus facilitating a better understanding of its infinite nature and structure.