Formulating and Predicting Sequences: A Quadratic Approach to the Series 3, 8, 15, 24, 35

Discovering patterns and formulating sequences is a fundamental aspect of mathematics. This article explores a particular sequence—3, 8, 15, 24, 35—and provides a step-by-step approach using quadratic formulas to predict subsequent terms. We will delve into the mathematical reasoning behind the sequence and its formula, and provide clear, actionable instructions for readers.

Understanding the Sequence: 3, 8, 15, 24, 35

Let's start with the sequence 3, 8, 15, 24, 35. To identify a possible formula, we will first calculate the differences between consecutive terms and then examine the second-order differences to determine the nature of the sequence.

Initial Differences:

First, let's calculate the differences between consecutive terms:

8 - 3 5 15 - 8 7 24 - 15 9 35 - 24 11

The first-order differences are: 5, 7, 9, 11.

Second-Order Differences:

Next, let's calculate the second-order differences:

7 - 5 2 9 - 7 2 11 - 9 2

The second-order differences are constant at 2, indicating that the sequence follows a quadratic formula.

Formulating the Quadratic Sequence

Given that the sequence follows a quadratic formula, we can express it as:

an An2 Bn C

To determine the coefficients A, B, C, we will use the initial terms of the sequence to form a system of equations.

Solving for Coefficients:

Using the first three terms of the sequence:

For n 1: A(1)2 B(1) C 3 → A B C 3 For n 2: A(2)2 B(2) C 8 → 4A 2B C 8 For n 3: A(3)2 B(3) C 15 → 9A 3B C 15

Substituting and solving these equations, we get:

1. A B C 3

2. 4A 2B C 8

3. 9A 3B C 15

Subtract the first equation from the second:

3A B 5 → Equation 4

Subtract the second equation from the third:

5A B 7 → Equation 5

Subtracting Equation 4 from Equation 5:

2A 2 → A 1

Substitute A 1 into Equation 4:

3(1) B 5 → B 2

Substitute A 1 and B 2 into Equation 1:

1 2 C 3 → C 0

Therefore, the quadratic formula for the sequence is:

an n2 2n

Predicting the Next Terms

Now, let's use the formula to predict the next terms in the sequence:

Next Term for n 6:

a6 62 2(6) 36 12 48

Next three terms:

For n 7: t7 72 2(7) 49 14 63 For n 8: t8 82 2(8) 64 16 80 For n 9: t9 92 2(9) 81 18 99

The sequence continues as: 3, 8, 15, 24, 35, 48, 63, 80, 99, ...

By following these steps, we can accurately predict future terms in a quadratic sequence, enhancing our understanding of numerical patterns.