Quadratic Sequence

A quadratic sequence is a sequence of numbers where the second differences are constant.

• First differences: Obtained by subtracting each term from the one that follows it.
• Second differences: Obtained by subtracting each first difference from the one that follows it.
• If the second differences are the same, it’s a quadratic sequence

The defining property of a quadratic sequence is that its second differences are constant and non-zero.

Example: Consider the sequence 5, 11, 19, 29, 41, …

Terms: 5 11 19 29 41

First differences: 6 8 10 12

Second differences: 2 2 2

The second differences are all 2, confirming that this is a quadratic sequence.

Some Real-Life examples are:

1. Electron Shells

The number of electrons in each energy level follows the formula 2n².
Example: 2, 8, 18, 32...

2. Motion and Gravity

When an object is dropped or thrown, its motion follows a quadratic pattern due to gravity.
Example: Distance covered over time forms a quadratic sequence.

3. Geometry and Construction

Area of squares follows a quadratic pattern.
Example: 1,4,9,16...1, 4, 9, 16...1,4,9,16...

General Formula

The n^th term of any quadratic sequence is expressed as:

T(n) = an² + bn + c

• a - rate of change of differences; a ≠ 0.
• b - rate of change of terms.
• c - constant term.
• n - position of the term (starting from n = 1).

Generating Terms of a Quadratic Sequence

Terms are generated by substituting n = 1, 2, 3, 4, 5, … into T(n) = an² + bn + c. Each substitution provides the value of the term at that position.

Example: Determine the first 5 terms for T(n) = 2n² − 3n + 5.

Substituting n = 1 through 5:

n	Calculation	T(n)
1	2(1) − 3(1) + 5	4
2	2(4) − 3(2) + 5	7
3	2(9) − 3(3) + 5	14
4	2(16) − 3(4) + 5	25
5	2(25) − 3(5) + 5	40

So, the terms are 4, 7, 14, 25, 40.

Finding the n^th Term of a Quadratic Sequence

Given only the terms of a quadratic sequence, the goal is to reverse-engineer the formula T(n) = an² + bn + c by determining the values of a, b, and c. This is done through a systematic use of the sequence's differences, using the following:

• a = (second difference) / 2
• b and c are solved using simultaneous equations formed from T(1) and T(2)

Steps to find the nth term:

Step 1: Compute the first differences by subtracting consecutive terms.

Step 2: Compute the second differences by subtracting consecutive first differences.

Step 3: Determine a use:

a = (constant second difference) / 2

Step 4: Substitute a into T(1) = a + b + c and T(2) = 4a + 2b + c to get two equations in b and c.

Step 5: Solve the simultaneous equations to find b and c.

Step 6: Write the complete formula T(n) = an² + bn + c and verify it using a term from the original sequence.

Example: Find the nth term for the given sequence: 3, 8, 15, 24, 35, ….

Step 1: Find the first differences.

Terms: 3 8 15 24 35
First differences: 5 7 9 11

Step 2: Find the second differences.

First differences: 5 7 9 11
Second differences: 2 2 2

Step 3: Find a

a = 2 / 2 = 1

So T(n) = n² + bn + c.

Step 4: Set up equations using T(1) and T(2).

T(1) = 1 + b + c = 3 → b + c = 2 … (i)
T(2) = 4 + 2b + c = 8 → 2b + c = 4 … (ii)

Step 5: Solve T(1) and T(2) simultaneously.

Subtracting (i) from (ii):

(2b + c) − (b + c) = 4 − 2
b = 2

Substituting back into (i):

2 + c = 2 → c = 0

So the sequence T(n) = n² + 2n.

Solved Examples

Example 1: In the quadratic sequence 5, __, 17, 26, 37, …, find the missing second term and the nth term formula.

Solution:

Let T(2) = k, so the first differences are (k − 5), (17 − k), 9, 11.

Since second differences in a quadratic sequence must be constant, consecutive second differences are equal:

(17 − k) − (k − 5) = 9 − (17 − k)

= 22 − 2k = k − 8

= 3k = 30

=> k = 10

With the sequence now complete as 5, 10, 17, 26, 37, the constant second difference = 2

Now, a = second difference /2 = 1.

So T(n) = n² + bn + c

Setting up equations from the first two terms:

T(1): 1 + b + c = 5 → b + c = 4 … (i)

T(2): 4 + 2b + c = 10 → 2b + c = 6 … (ii)

Subtracting (i) from (ii): b = 2

Substituting into (i): c = 2

So, T(n) = n² + 2n + 2.

Example 2: The nth term of a quadratic sequence is T(n) = 2n² − n.

• Find the 9th term.
• Which term of the sequence has the value 66?

Solution:

(a) For 9th term, substituting n = 9 directly into the formula:

T(9) = 2(9²) − 9

T(9) = 2(81) − 9

T(9) = 162 − 9 = 153.

(b) Setting T(n) = 66 and rearranging into standard quadratic form:

2n² − n − 66 = 0

Applying the quadratic formula with A = 2, B = −1, C = −66:

Discriminant = (−1)² − 4(2)(−66) = 1 + 528 = 529

-> √529 = 23

n = (1 + 23) / 4 = 6

(The root n = (1 − 23)/4 = −5.5 is rejected since n must be a positive integer.)

Practice Problems

Problem 1: Determine whether the sequence 1, 5, 11, 19, 29, … is a quadratic sequence. Justify your answer using differences.

Problem 2: The nth term of a sequence is given by T(n) = 3n² − 2n + 4. Write down the first 5 terms of the sequence.

Problem 3: Find the nth term formula for the quadratic sequence: 2, 10, 22, 38, 58, …

Problem 4: The nth term of a quadratic sequence is T(n) = n² + 4n − 2. Find the value of n for which T(n) = 43.

Problem 5: A quadratic sequence has its first three terms as 5, 14, and 27. The nth term can be written in the form an² + bn + c. Find the value of T(20).