Problem 57 - College Algebra with Corequisite Support

Question

Find the first five terms of the sequence \(a_{1}=\frac{87}{111}\), \(a_{n}=\frac{4}{3} a_{n-1}+\frac{12}{37} .\) Use the \(>\) Frac feature to give fractional results.

Step-by-Step Solution

Answer

The first five terms of the sequence are \( \frac{29}{37}, \frac{1680}{1333}, \frac{2473}{1233}, \frac{2993}{999}, \frac{4335}{999} \).

Step 1: Find the First Term

The problem states that the first term of the sequence, \( a_1 \), is given. Therefore, \( a_1 = \frac{87}{111} \). We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 3. This gives us: \( a_1 = \frac{29}{37} \).

Step 2: Calculate the Second Term

To find the second term \( a_2 \), use the recurrence relation \( a_n = \frac{4}{3} a_{n-1} + \frac{12}{37} \). Substitute \( n = 2 \) and \( a_1 = \frac{29}{37} \):\[ a_2 = \frac{4}{3} \times \frac{29}{37} + \frac{12}{37} \]Calculate \( \frac{4}{3} \times \frac{29}{37} = \frac{116}{111} \) and then find a common denominator for the two fractions:\[ a_2 = \frac{116}{111} \times \frac{3}{3} + \frac{12}{37} \times \frac{111}{111} = \frac{348}{333} + \frac{1332}{1333} \]Simplifying gives \( a_2 = \frac{348 + 1332}{1333} = \frac{1680}{1333} \)

Step 3: Calculate the Third Term

Using \( a_3 = \frac{4}{3} a_2 + \frac{12}{37} \), substitute \( a_2 = \frac{1680}{1333} \):\[ a_3 = \frac{4}{3} \times \frac{1680}{1333} + \frac{12}{37} \]Calculate \( \frac{4}{3} \times \frac{1680}{1333} = \frac{6720}{3999} \), then find a common denominator:\[ a_3 = \frac{6720}{3999} \times \frac{37}{37} + \frac{12}{37} \times \frac{3999}{3999} = \frac{248640}{147963} + \frac{47988}{147963} \]Simplify to get \( a_3 = \frac{296628}{147963} = \frac{2473}{1233} \).

Step 4: Calculate the Fourth Term

To find \( a_4 \) using \( a_4 = \frac{4}{3} a_3 + \frac{12}{37} \), substitute \( a_3 = \frac{2473}{1233} \):\[ a_4 = \frac{4}{3} \times \frac{2473}{1233} + \frac{12}{37} \]Calculate \( \frac{4}{3} \times \frac{2473}{1233} = \frac{9892}{3699} \). Find a common denominator:\[ a_4 = \frac{9892}{3699} \times \frac{37}{37} + \frac{12}{37} \times \frac{3699}{3699} = \frac{365004}{136863} + \frac{44388}{136863} \]Simplify to get \( a_4 = \frac{409392}{136863} = \frac{2993}{999} \).

Step 5: Calculate the Fifth Term

Use the formula \( a_5 = \frac{4}{3} a_4 + \frac{12}{37} \), where \( a_4 = \frac{2993}{999} \):\[ a_5 = \frac{4}{3} \times \frac{2993}{999} + \frac{12}{37} \]Calculate \( \frac{4}{3} \times \frac{2993}{999} = \frac{11972}{2997} \). Convert both fractions to a common denominator:\[ a_5 = \frac{11972}{2997} \times \frac{37}{37} + \frac{12}{37} \times \frac{2997}{2997} = \frac{443964}{110889} + \frac{35964}{110889} \]Simplify to get \( a_5 = \frac{479928}{110889} = \frac{4335}{999} \).

Key Concepts

Recurrence Relations

A recurrence relation is a way to define a sequence by using previous terms. It's like building a story where each chapter is based on the events of the previous chapter. In this exercise, the relation given is: \( a_n = \frac{4}{3} a_{n-1} + \frac{12}{37} \).
This means that to find any term \(a_n\), we use the term before it, \(a_{n-1}\), and apply the relation to calculate the next term. With this setup, the initial term, known as \(a_1\), is crucial because it's our starting point. In simple terms, recurrence relations turn the process of finding terms in a sequence into a step-by-step journey from one term to the next.
By understanding how recurrence relations work, any term in an infinite sequence can be precisely calculated, as long as we know how to start and follow the rule given.

Fraction Simplification

Fraction simplification makes calculations easier and results neater. Simplification involves reducing the numerator and denominator of a fraction to their smallest possible values by dividing them by their greatest common divisor (GCD).
In this exercise, the first term of the sequence, \( a_1 = \frac{87}{111} \), was simplified to \( \frac{29}{37} \) because both 87 and 111 can be divided by 3, which is their GCD. Simplifying fractions doesn’t change their value; it just writes them in a clearer form.
Simplifying fractions at every step when using recurrence relations often makes operations like addition, subtraction, or multiplication easier and helps in identifying patterns within sequences.

Sequence Terms Calculation

Calculating terms in a sequence can seem daunting, but by using the given recurrence relation, the process is simplified. First, identify the initial condition or first term, which is \( a_1 \). Then, proceed to calculate each subsequent term using the relation.
For example, if the relation is \( a_n = \frac{4}{3} a_{n-1} + \frac{12}{37} \), insert the last computed term (\(a_{n-1}\)) into the formula to obtain \(a_n\).
For the second term: Start with the first term: \( a_2 = \frac{4}{3} \times \frac{29}{37} + \frac{12}{37} \), then perform the necessary arithmetic operations. As shown in the steps, find a common denominator to effectively add two fractions.

• Multiply the fraction by a common factor to get like denominators.
• Once fractions are comparable, sum the numerators.
• Simplify the resulting fraction if needed.

This process is repeated with each subsequent term, ensuring accuracy every step of the way.