Quadratic Sequences – nth Term, Second Differences, and Patterns
1. Introduction
In this lesson, you will learn about quadratic sequences. Unlike arithmetic sequences where differences between consecutive terms are constant, quadratic sequences have differences that change at a constant rate. You will learn to identify these sequences and find their general n\text{th} term formulas.
2. Core Concept Explanation
A sequence is quadratic if the general term U_n (the n\text{th} term) can be written as:
Where a, b, c are constants and a \neq 0.
In an arithmetic sequence, the first differences (difference between consecutive terms) are constant. In a quadratic sequence, the first differences form their own arithmetic sequence, meaning the second differences (differences of the first differences) are constant.
The relationship between the second differences and the coefficient a is:
3. Step-by-Step Method
To find the n\text{th} term formula U_n = an^2 + bn + c of a quadratic sequence:
1. Find the first differences between consecutive terms.
2. Find the second differences. Confirm they are constant.
3. Determine a by dividing the constant second difference by 2 (a = \text{Second Difference} / 2).
4. Subtract an^2 from each term of the original sequence. The resulting values will form a simple linear (arithmetic) sequence: bn + c.
5. Find the linear formula for this new sequence to determine the values of b and c.
6. Combine the parts to write the final formula: U_n = an^2 + bn + c.
4. Worked Examples
* Example 1: Find the n\text{th} term of the sequence: 3, 8, 15, 24, 35, \dots
* Solution:
1. Terms: U_1=3, U_2=8, U_3=15, U_4=24, U_5=35.
2. First differences: 5, 7, 9, 11.
3. Second differences: 2, 2, 2. Constant is 2.
4. Since 2a = 2, we get a = 1.
5. Subtract an^2 = n^2 from the sequence:
* For n=1: 3 - 1^2 = 2
* For n=2: 8 - 2^2 = 4
* For n=3: 15 - 3^2 = 6
* For n=4: 24 - 4^2 = 8
The new sequence is 2, 4, 6, 8, \dots, which has the formula 2n.
6. Thus, b = 2 and c = 0.
7. Final formula: U_n = n^2 + 2n.
* Example 2: Find the n\text{th} term of the sequence: 4, 9, 18, 31, 48, \dots
* Solution:
1. First differences: 5, 9, 13, 17.
2. Second differences: 4, 4, 4.
3. Since 2a = 4, we get a = 2.
4. Subtract 2n^2 from each term:
* n=1: 4 - 2(1) = 2
* n=2: 9 - 2(4) = 1
* n=3: 18 - 2(9) = 0
* n=4: 31 - 2(16) = -1
The new sequence is 2, 1, 0, -1, \dots
5. Find the linear term: first term is 2, common difference is -1. So U_{\text{linear}} = -n + 3.
6. Combine: U_n = 2n^2 - n + 3.
* Example 3: Calculate the 10th term of the sequence defined by U_n = 3n^2 + 2n - 1.
* Solution:
Substitute n = 10:
5. Exam Relevance
* Example WAEC Question: Find the formula for the n\text{th} term of the sequence: 2, 6, 12, 20, 30, \dots
* Solution:
First differences: 4, 6, 8, 10.
Second differences: 2, 2, 2.
Here, 2a = 2 \Rightarrow a = 1.
Subtract n^2:
* n=1: 2 - 1 = 1, n=2: 6 - 4 = 2, n=3: 12 - 9 = 3.
The remaining sequence is 1, 2, 3, \dots which is simply n.
The formula is U_n = n^2 + n.
6. Common Mistakes
* Confusing 2a with a: Setting a equal to the second difference directly instead of dividing it by 2.
* Incorrect n index: Starting the sequence calculations with n=0 instead of n=1.
7. Practice Questions
1. Find the n\text{th} term of the sequence 5, 12, 23, 38, 57, \dots
* Answer: U_n = 2n^2 + n + 2.
2. What is the 5th term of the sequence U_n = -n^2 + 5n?
* Answer: U_5 = 0.
8. Summary / Cheat Sheet
* General form: U_n = an^2 + bn + c.
* Key formula: 2a = \text{Second Difference}.
* The second differences must be constant for a sequence to be quadratic.
