Problem 68 - College Algebra

Question

What term in the sequence \(a_{n}=\frac{n^{2}+4 n+4}{2(n+2)}\) has the value 41 ? Verify the result.

Step-by-Step Solution

Answer

The 80th term of the sequence equals 41.

Step 1: Identifying the Term with Value 41

We need to find the term in the sequence that equals 41, so we set the expression for the sequence equal to 41 and solve: \[\frac{n^2 + 4n + 4}{2(n+2)} = 41.\]

Step 2: Multiply Through by the Denominator

Multiply both sides by the denominator to eliminate the fraction:\[n^2 + 4n + 4 = 41 imes (2n + 4).\]Expanding the right side, we have:\[n^2 + 4n + 4 = 82n + 164.\]

Step 3: Rearrange the Equation

Rearrange the equation to bring all terms to one side so you can solve for \(n\):\[n^2 + 4n + 4 - 82n - 164 = 0.\]This simplifies to:\[n^2 - 78n - 160 = 0.\]

Step 4: Solve the Quadratic Equation

Use the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to solve the equation. Here, \(a = 1\), \(b = -78\), \(c = -160\):\[n = \frac{-(-78) \pm \sqrt{(-78)^2 - 4 \cdot 1 \cdot (-160)}}{2 \cdot 1}.\]Calculate the discriminant:\[n = \frac{78 \pm \sqrt{6084 + 640}}{2} = \frac{78 \pm \sqrt{6724}}{2}.\]Since \(\sqrt{6724} = 82\), we have two potential solutions:\[n = \frac{78 + 82}{2} \quad \text{and} \quad n = \frac{78 - 82}{2}.\]

Step 5: Evaluate the Two Solutions

The potential solutions are:\[n = \frac{160}{2} = 80 \quad \text{and} \quad n = \frac{-4}{2} = -2.\]Since \(n\) must be a positive integer, we accept \(n = 80\).

Step 6: Verification Step

Substitute \(n = 80\) back into the sequence formula to verify:\[a_{80} = \frac{80^2 + 4(80) + 4}{2(80+2)} = \frac{6400 + 320 + 4}{164} = \frac{6724}{164} = 41.\]Thus, the term at \(n = 80\) is indeed 41.

Key Concepts

Algebraic Sequences

An algebraic sequence is a list of numbers where each term is formed using a specific algebraic expression. In our exercise, the sequence is defined by the formula: \[a_{n} = \frac{n^{2} + 4n + 4}{2(n+2)}.\] This means that for any positive integer value of \(n\), we can compute the term of the sequence by plugging \(n\) into the expression.

• The numerator, \(n^{2} + 4n + 4\), is a polynomial function describing the growth pattern of the sequence.
• The denominator, \(2(n+2)\), is a simple linear function that affects how the sequence progresses.

These sequences are essential in algebra as they allow us to explore number patterns and relationships. By providing an equation for the sequence, we can easily find any term without listing each one individually.

Quadratic Formula

The quadratic formula is a powerful tool used to find the solutions of quadratic equations, which are equations of the form \[ax^2 + bx + c = 0.\] For this exercise, we needed to find \(n\) that makes the sequence value equal to 41, leading us to the quadratic equation:\[n^2 - 78n - 160 = 0.\] Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we can solve for \(n\). Here, \(a = 1\), \(b = -78\), and \(c = -160\).

• First, calculate the discriminant: \(b^2 - 4ac\), which helps to determine the number and type of solutions.
• The discriminant in our equation was positive, resulting in two real solutions.

The quadratic formula gives us two potential values for \(n\). Identifying and evaluating these solutions is crucial to finding the correct term in our sequence.

Mathematical Verification

Verification in mathematics is crucial for confirming that a solution is correct. After solving for \(n\) and finding potential solutions, mathematical verification ensures that our results are consistent with the original problem.

• In our exercise, we found \(n = 80\) as the valid solution for our sequence.
• Verification involves substituting \(n = 80\) back into the sequence formula to confirm that it indeed yields 41:

\[a_{80} = \frac{80^2 + 4(80) + 4}{2(80+2)} = \frac{6724}{164} = 41.\] This verification step is important as it checks that all calculations and assumptions were correctly followed. Without it, errors might go unnoticed. Thus, a thorough mathematical verification builds confidence in the solution's accuracy.