Problem 19 - Thomas Calculus

Question

Find a formula for the \(n\)th term of the sequence. $$ 0,3,8,15,24, \dots $$

Step-by-Step Solution

Answer

The formula for the \(n\)th term is \(a_n = n^2 - 1\).

Step 1: Identify the Pattern

Let's examine the differences between consecutive terms of the sequence: \(3 - 0 = 3\), \(8 - 3 = 5\), \(15 - 8 = 7\), \(24 - 15 = 9\). The differences are: 3, 5, 7, 9, which increase by 2 each time. This suggests a quadratic pattern.

Step 2: Assume a Quadratic Formula

Since the differences increase linearly, assume the sequence follows a quadratic formula of the form \(a_n = an^2 + bn + c\).

Step 3: Set Up Equations Using Given Terms

Use the known terms of the sequence to set up equations. For the first few terms, substitute \(n\) with 1, 2, and 3:1. \(a(1)^2 + b(1) + c = 0\)2. \(a(2)^2 + b(2) + c = 3\)3. \(a(3)^2 + b(3) + c = 8\)

Step 4: Solve the System of Equations

We solve the equations from Step 3:1. \(a + b + c = 0\)2. \(4a + 2b + c = 3\)3. \(9a + 3b + c = 8\)Subtract the first equation from the second and third equations:- (2) - (1): \(3a + b = 3\)- (3) - (2): \(5a + b = 5\)Solve this system:- \(3a + b = 3\)- \(5a + b = 5\)Subtracting these gives \(2a = 2\), so \(a = 1\). Substitute \(a = 1\) into \(3a + b = 3\) to find \(b\):- \(3(1) + b = 3\)- \(b = 0\)Finally, substitute \(a = 1\) and \(b = 0\) into \(a + b + c = 0\) to find \(c\):- \(1 + 0 + c = 0\)- \(c = -1\)

Step 5: Form the General Formula

Using the above values of \(a\), \(b\), and \(c\), the formula for the \(n\)th term is: \(a_n = n^2 - 1\).

Key Concepts

Sequence Pattern

Understanding the sequence pattern is the first step towards identifying the right formula for a sequence. A sequence pattern defines how the numbers in a sequence progress from one term to the next. In our sequence, the terms are 0, 3, 8, 15, 24. To unearth the pattern, we need to look at the differences between consecutive terms.

• The difference between the first two terms, 3 and 0, is 3.
• Next, 8 minus 3 gives us 5.
• Then, 15 minus 8 results in 7.
• Finally, 24 minus 15 equals 9.

These differences form another sequence: 3, 5, 7, 9. Notice how each successive difference increases by 2? This increasing pattern suggests that the original sequence is quadratic. Recognizing this pattern is crucial because it hints at the form the formula should take. It's a valuable step that guides us to explore quadratic sequences, which typically possess a second-order difference that is constant.

nth Term Formula

A key part of working with sequences is determining the nth term formula, which allows you to find any term in the sequence without having to list all the preceding ones. For quadratic sequences, the nth term is generally expressed as \(a_n = an^2 + bn + c\). Here, \(a\), \(b\), and \(c\) are constants that we need to find.
By examining the quadratic nature of a sequence, we can find these constants through known terms. In our example, the quadratic expression should satisfy the conditions set by the terms of the sequence. We'll substitute values into different forms of equations using the sequence's terms, which allows us to solve for \(a\), \(b\), and \(c\). Once we have these constants, we obtain the formula for any term, which is useful in predicting future terms or analyzing the sequence further.

Difference Method

The difference method is instrumental in identifying whether a sequence is quadratic and finding the coefficients for the nth term formula. By calculating the differences between consecutive terms, as we did with 3, 5, 7, and 9, we get the first layer of information. Then, by observing that this set of differences increases by a constant amount (2 in our case), we conclude that the sequence is quadratic.
For quadratic sequences, the second difference (the difference of differences) is constant. This method points us directly to a quadratic formula, indicating that \(a_n\) can be described as a polynomial of degree 2. Knowing this sequence type helps us construct the nth term formula effectively.

• This insight reduces complexity when dealing with sequences, as knowing the type (linear, quadratic, etc.) narrows down the potential formulas to test.
• The method is particularly useful because it provides a straightforward path to set up simultaneous equations, which are crucial for determining the exact values of \(a\), \(b\), and \(c\).

Solving Equations

Once you've set up your simultaneous equations from the known terms of the sequence, the next step is solving them to find \(a\), \(b\), and \(c\). Simultaneous equations are simply a set of equations containing multiple variables. For the quadratic sequence in our example, we have three equations:

• \(a + b + c = 0\)
• \(4a + 2b + c = 3\)
• \(9a + 3b + c = 8\)

Subtract one equation from another to eliminate a variable and solve step-by-step. Begin by eliminating \(c\) from the equations, arriving at simpler ones like \(3a + b = 3\) and \(5a + b = 5\). Further simplify by subtracting these to find that \(2a = 2\), leading to \(a = 1\). Then backtrack to find \(b\) and \(c\).
This approach of stepping through equations methodically helps in efficiently finding the values needed to complete your nth term formula. It teaches the importance of careful setup and simplification of various algebraic expressions, crucial skills in algebra.