0.3 Quadratic sequences

Siyavula textbooks: grade 11 Page 1 / 1

Introduction

In Grade 10, you learned about arithmetic sequences, where the difference between consecutive terms was constant. In this chapter we learn about quadratic sequences.

What is a quadratic sequence ?

Quadratic Sequence

A quadratic sequence is a sequence of numbers in which the second differences between each consecutive term differ by the same amount, called a common second difference.

For example,

1; 2; 4; 7; 11; ...

is a quadratic sequence. Let us see why ...

If we take the difference between consecutive terms, then:

\begin{matrix} a_{2} - a_{1} & = 2 - 1 & = 1 \\ a_{3} - a_{2} & = 4 - 2 & = 2 \\ a_{4} - a_{3} & = 7 - 4 & = 3 \\ a_{5} - a_{4} & = 11 - 7 & = 4 \end{matrix}

We then work out the second differences , which is simply obtained by taking the difference between the consecutive differences { $1; 2; 3; 4; ...$ } obtained above:

\begin{matrix} 2 - 1 & = & 1 \\ 3 - 2 & = & 1 \\ 4 - 3 & = & 1 \\ ... \end{matrix}

We then see that the second differences are equal to 1. Thus, [link] is a quadratic sequence .

Note that the differences between consecutive terms (that is, the first differences) of a quadratic sequence form a sequence where there is a constant difference between consecutive terms. In the above example, the sequence of { $1; 2; 3; 4; ...$ }, which is formed by taking the differences between consecutive terms of [link] , has a linear formula of the kind $a x + b$ .

Quadratic sequences

The following are also examples of quadratic sequences:

\begin{matrix} 3; 6; 10; 15; 21; ... \\ 4; 9; 16; 25; 36; ... \\ 7; 17; 31; 49; 71; ... \\ 2; 10; 26; 50; 82; ... \\ 31; 30; 27; 22; 15; ... \end{matrix}

Can you calculate the common second difference for each of the above examples?

Write down the next two terms and find a formula for the $n^{th}$ term of the sequence $5, 12, 23, 38, . . ., . . .,$

Find the first differences between the terms.
i.e. $7, 11, 15$
Find the second differences between the terms.
the second difference is 4.

So continuing the sequence, the differences between each term will be:

$15 + 4 = 19$

$19 + 4 = 23$
Finding the next two terms.
So the next two terms in the sequence willl be:

$38 + 19 = 57$

$57 + 23 = 80$

So the sequence will be: $5, 12, 23, 38, 57, 80$
We now need to find the formula for this sequence.
We know that the second difference is 4. The start of the formula will therefore be $2 n^{2}$ .
We now need to work out the next part of the sequence.
If $n = 1$ , you have to get the value of term one, which is 5 in this particular sequence. The difference between $2 n^{2} = 2$ and original number (5) is 3, which leads to $n + 2$ .

Check if it works for the second term, i.e. when $n = 2$ .

Then $2 n^{2} = 8$ . The difference between term two (12) and 8 is 4, which is can be written as $n + 2$ .

So for the sequence $5, 12, 23, 38, . . .$ the formula for the $n^{th}$ term is $2 n^{2} + n + 2$ .

General case

If the sequence is quadratic, the $n^{th}$ term should be $T_{n} = a n^{2} + b n + c$

TERMS	$a + b + c$		$4 a + 2 b + c$		$9 a + 3 b + c$
$1^{st}$ difference		$3 a + b$		$5 a + b$		$7 a + b$
$2^{nd}$ difference			$2 a$		$2 a$

In each case, the second difference is $2 a$ . This fact can be used to find $a$ , then $b$ then $c$ .

The following sequence is quadratic: $8, 22, 42, 68, . . .$ Find the rule.

Assume that the rule is $a n^{2} + b n + c$

TERMS 8 22 42 68

$1^{st}$ difference 14 20 26

$2^{nd}$ difference 6 6 6
Determine values for $a, b$ and $c$
$\begin{matrix} Then 2 a & = & 6 \\ which gives a & = & 3 \\ And 3 a + b & = & 14 \\ ∴ 9 + b & = & 14 \\ b & = & 5 \\ And a + b + c & = & 8 \\ ∴ 3 + 5 + c & = & 8 \\ c & = & 0 \end{matrix}$
Find the rule
The rule is therefore: $n^{th} t e r m = 3 n^{2} + 5 n$
Check answer
For

$\begin{matrix} n & = & 1, T_{1} = 3 {(1)}^{2} + 5 (1) = 8 \\ n & = & 2, T_{2} = 3 {(2)}^{2} + 5 (2) = 22 \\ n & = & 3, T_{3} = 3 {(3)}^{2} + 5 (3) = 42 \end{matrix}$

Derivation of the $n^{th}$ -term of a quadratic sequence

Let the $n^{t h}$ -term for a quadratic sequence be given by

\begin{matrix} a_{n} = A \cdot n^{2} + B \cdot n + C \end{matrix}

where $A$ , $B$ and $C$ are some constants to be determined.

\begin{matrix} a_{n} & = & A \cdot n^{2} + B \cdot n + C \\ a_{1} & = & A {(1)}^{2} + B (1) + C = A + B + C \\ a_{2} & = & A {(2)}^{2} + B (2) + C = 4 A + 2 B + C \\ a_{3} & = & A {(3)}^{2} + B (3) + C = 9 A + 3 B + C \end{matrix}

\begin{matrix} Let d & \equiv & a_{2} - a_{1} \\ ∴ d & = & 3 A + B \end{matrix}

\Rightarrow B = d - 3 A

The common second difference is obtained from

\begin{matrix} D & = & (a_{3} - a_{2}) - (a_{2} - a_{1}) \\ = & (5 A + B) - (3 A + B) \\ = & 2 A \end{matrix}

\Rightarrow A = \frac{D}{2}

Therefore, from [link] ,

B = d - \frac{3}{2} \cdot D

From [link] ,

C = a_{1} - (A + B) = a_{1} - \frac{D}{2} - d + \frac{3}{2} \cdot D

∴ C = a_{1} + D - d

Finally, the general equation for the $n^{t h}$ -term of a quadratic sequence is given by

a_{n} = \frac{D}{2} \cdot n^{2} + (d - \frac{3}{2} D) \cdot n + (a_{1} - d + D)

Study the following pattern: 1; 7; 19; 37; 61; ...

What is the next number in the sequence ?
Use variables to write an algebraic statement to generalise the pattern.
What will the 100th term of the sequence be ?

The next number in the sequence
The numbers go up in multiples of 6

$1 + 6 (1) = 7$ , then $7 + 6 (2) = 19$

$19 + 6 (3) = 37$ , then $37 + 6 (4) = 61$

Therefore $61 + 6 (5) = 91$

The next number in the sequence is 91.
Generalising the pattern

TERMS 1 7 19 37 61

$1^{st}$ difference 6 12 18 24

$2^{nd}$ difference 6 6 6 6

The pattern will yield a quadratic equation since the second difference is constant

Therefore $a n^{2} + b n + c = y$

For the first term: $n = 1$ , then $y = 1$

For the second term: $n = 2$ , then $y = 7$

For the third term: $n = 3$ , then $y = 19$

etc....
Setting up sets of equations
$\begin{matrix} a + b + c & = & 1 \\ 4 a + 2 b + c & = & 7 \\ 9 a + 3 b + c & = & 19 \end{matrix}$
Solve the sets of equations
$\begin{matrix} eqn (2) - eqn (1) : & 3 a + b = 6 \\ eqn (3) - eqn (2) : & 5 a + b = 12 \\ eqn (5) - eqn (4) : & 2 a = 6 \\ ∴ & a = 3, b = - 3 a n d c = 1 \end{matrix}$
Final answer
The general formula for the pattern is $3 n^{2} - 3 n + 1$
Term 100
Substitute n with 100:

$3 {(100)}^{2} - 3 (100) + 1 = 29 701$

The value for term 100 is 29 701.

Plotting a graph of terms of a quadratic sequence

Plotting $a_{n}$ vs. $n$ for a quadratic sequence yields a parabolic graph.

Given the quadratic sequence,

\begin{matrix} 3; 6; 10; 15; 21; ... \end{matrix}

If we plot each of the terms vs. the corresponding index, we obtain a graph of a parabola.

End of chapter exercises

Find the first 5 terms of the quadratic sequence defined by:
$a_{n} = n^{2} + 2 n + 1$
Determine which of the following sequences is a quadratic sequence by calculating the common second difference:
1. $6; 9; 14; 21; 30; ...$
2. $1; 7; 17; 31; 49; ...$
3. $8; 17; 32; 53; 80; ...$
4. $9; 26; 51; 84; 125; ...$
5. $2; 20; 50; 92; 146; ...$
6. $5; 19; 41; 71; 109; ...$
7. $2; 6; 10; 14; 18; ...$
8. $3; 9; 15; 21; 27; ...$
9. $10; 24; 44; 70; 102; ...$
10. $1; 2, 5; 5; 8, 5; 13; ...$
11. $2, 5; 6; 10, 5; 16; 22, 5; ...$
12. $0, 5; 9; 20, 5; 35; 52, 5; ...$
Given $a_{n} = 2 n^{2}$ , find for which value of $n$ , $a_{n} = 242$
Given $a_{n} = {(n - 4)}^{2}$ , find for which value of $n$ , $a_{n} = 36$
Given $a_{n} = n^{2} + 4$ , find for which value of $n$ , $a_{n} = 85$
Given $a_{n} = 3 n^{2}$ , find $a_{11}$
Given $a_{n} = 7 n^{2} + 4 n$ , find $a_{9}$
Given $a_{n} = 4 n^{2} + 3 n - 1$ , find $a_{5}$
Given $a_{n} = 1, 5 n^{2}$ , find $a_{10}$
For each of the quadratic sequences, find the common second difference, the formula for the general term and then use the formula to find a 100 .
1. $4, 7, 12, 19, 28, ...$
2. $2, 8, 18, 32, 50, ...$
3. $7, 13, 23, 37, 55, ...$
4. $5, 14, 29, 50, 77, ...$
5. $7, 22, 47, 82, 127, ...$
6. $3, 10, 21, 36, 55, ...$
7. $3, 7, 13, 21, 31, ...$
8. $3, 9, 17, 27, 39, ...$

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Source: OpenStax, Siyavula textbooks: grade 11 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11243/1.3

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