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This module is included inLens: Siyavula: Mathematics (Gr. 10-12)
By: Siyavula

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## 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?

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 ; ... 1 ; 2 ; 4 ; 7 ; 11 ; ...
(1)

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

If we take the difference between consecutive terms, then:

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 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
(2)

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

2 - 1 = 1 3 - 2 = 1 4 - 3 = 1 ... 2 - 1 = 1 3 - 2 = 1 4 - 3 = 1 ...
(3)

We then see that the second differences are equal to 1. Thus, Equation 1 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;...1;2;3;4;...}, which is formed by taking the differences between consecutive terms of Equation 1, has a linear formula of the kind ax+bax+b.

The following are also examples of quadratic sequences:

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 ; ... 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 ; ...
(4)

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 n th term of the sequence 5,12,23,38,...,...,5,12,23,38,...,...,

#### Solution

1. Step 1. Find the first differences between the terms. :

i.e. 7,11,157,11,15

2. Step 2. 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 15 + 4 = 19

19 + 4 = 23 19 + 4 = 23

3. Step 3. Finding the next two terms. :

So the next two terms in the sequence willl be:

38 + 19 = 57 38 + 19 = 57

57 + 23 = 80 57 + 23 = 80

So the sequence will be: 5,12,23,38,57,805,12,23,38,57,80

4. Step 4. 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 2n22n2.

5. Step 5. We now need to work out the next part of the sequence. :

If n=1n=1, you have to get the value of term one, which is 5 in this particular sequence. The difference between 2n2=22n2=2 and original number (5) is 3, which leads to n+2n+2.

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

Then 2n2=82n2=8. The difference between term two (12) and 8 is 4, which is can be written as n+2n+2.

So for the sequence 5,12,23,38,...5,12,23,38,... the formula for the n th n th term is 2n2+n+22n2+n+2.

### General Case

If the sequence is quadratic, the n th n th term should be Tn=an2+bn+cTn=an2+bn+c

 TERMS a + b + c a + b + c 4 a + 2 b + c 4 a + 2 b + c 9 a + 3 b + c 9 a + 3 b + c 1st1st difference 3 a + b 3 a + b 5 a + b 5 a + b 7 a + b 7 a + b 2nd2nd difference 2 a 2 a 2 a 2 a

In each case, the second difference is 2a2a. This fact can be used to find aa, then bb then cc.

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

##### Solution
1. Step 1. Assume that the rule is an2+bn+can2+bn+c :
 TERMS 8 22 42 68 1st1st difference 14 20 26 2nd2nd difference 6 6 6
2. Step 2. Determine values for a,ba,b and cc :
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 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
(5)
3. Step 3. Find the rule :

The rule is therefore:    n th term=3n2+5nn th term=3n2+5n

4. Step 4. Check answer :

For

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 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
(6)

#### Derivation of the n th n th -term of a Quadratic Sequence

Let the nthnth-term for a quadratic sequence be given by

a n = A · n 2 + B · n + C a n = A · n 2 + B · n + C
(7)

where AA, BB and CC are some constants to be determined.

a n = A · n 2 + B · 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 a n = A · n 2 + B · 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
(8)
Let d a 2 - a 1 d = 3 A + B Let d a 2 - a 1 d = 3 A + B
(9)
B = d - 3 A B = d - 3 A
(10)

The common second difference is obtained from

D = ( a 3 - a 2 ) - ( a 2 - a 1 ) = ( 5 A + B ) - ( 3 A + B ) = 2 A D = ( a 3 - a 2 ) - ( a 2 - a 1 ) = ( 5 A + B ) - ( 3 A + B ) = 2 A
(11)
A = D 2 A = D 2
(12)

Therefore, from Equation 10,

B = d - 3 2 · D B = d - 3 2 · D
(13)

From Equation 8,

C = a 1 - ( A + B ) = a 1 - D 2 - d + 3 2 · D C = a 1 - ( A + B ) = a 1 - D 2 - d + 3 2 · D
(14)
C = a 1 + D - d C = a 1 + D - d
(15)

Finally, the general equation for the nthnth-term of a quadratic sequence is given by

a n = D 2 · n 2 + ( d - 3 2 D ) · n + ( a 1 - d + D ) a n = D 2 · n 2 + ( d - 3 2 D ) · n + ( a 1 - d + D )
(16)

#### Exercise 3: Using a set of equations

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

1. What is the next number in the sequence ?
2. Use variables to write an algebraic statement to generalise the pattern.
3. What will the 100th term of the sequence be ?
##### Solution
1. Step 1. The next number in the sequence :

The numbers go up in multiples of 6

1+6(1)=71+6(1)=7, then 7+6(2)=197+6(2)=19

19+6(3)=3719+6(3)=37, then 37+6(4)=6137+6(4)=61

Therefore 61+6(5)=9161+6(5)=91

The next number in the sequence is 91.

2. Step 2. Generalising the pattern :
 TERMS 1 7 19 37 61 1st1st difference 6 12 18 24 2nd2nd difference 6 6 6 6

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

Therefore an2+bn+c=yan2+bn+c=y

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

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

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

etc....

3. Step 3. Setting up sets of equations :
a + b + c = 1 4 a + 2 b + c = 7 9 a + 3 b + c = 19 a + b + c = 1 4 a + 2 b + c = 7 9 a + 3 b + c = 19
(17)
4. Step 4. Solve the sets of equations :
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 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
(18)
5. Step 5. Final answer :

The general formula for the pattern is 3n2-3n+13n2-3n+1

6. Step 6. Term 100 :

Substitute n with 100:

3 ( 100 ) 2 - 3 ( 100 ) + 1 = 29 701 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 anan vs. nn for a quadratic sequence yields a parabolic graph.

3 ; 6 ; 10 ; 15 ; 21 ; ... 3 ; 6 ; 10 ; 15 ; 21 ; ...
(19)

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

## End of chapter Exercises

1. Find the first 5 terms of the quadratic sequence defined by:
an=n2+2n+1an=n2+2n+1
(20)
2. Determine which of the following sequences is a quadratic sequence by calculating the common second difference:
1. 6;9;14;21;30;...6;9;14;21;30;...
2. 1;7;17;31;49;...1;7;17;31;49;...
3. 8;17;32;53;80;...8;17;32;53;80;...
4. 9;26;51;84;125;...9;26;51;84;125;...
5. 2;20;50;92;146;...2;20;50;92;146;...
6. 5;19;41;71;109;...5;19;41;71;109;...
7. 2;6;10;14;18;...2;6;10;14;18;...
8. 3;9;15;21;27;...3;9;15;21;27;...
9. 10;24;44;70;102;...10;24;44;70;102;...
10. 1;2,5;5;8,5;13;...1;2,5;5;8,5;13;...
11. 2,5;6;10,5;16;22,5;...2,5;6;10,5;16;22,5;...
12. 0,5;9;20,5;35;52,5;...0,5;9;20,5;35;52,5;...
3. Given an=2n2an=2n2, find for which value of nn, an=242an=242
4. Given an=(n-4)2an=(n-4)2, find for which value of nn, an=36an=36
5. Given an=n2+4an=n2+4, find for which value of nn, an=85an=85
6. Given an=3n2an=3n2, find a11a11
7. Given an=7n2+4nan=7n2+4n, find a9a9
8. Given an=4n2+3n-1an=4n2+3n-1, find a5a5
9. Given an=1,5n2an=1,5n2, find a10a10
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 a100a100.
1. 4,7,12,19,28,...4,7,12,19,28,...
2. 2,8,18,32,50,...2,8,18,32,50,...
3. 7,13,23,37,55,...7,13,23,37,55,...
4. 5,14,29,50,77,...5,14,29,50,77,...
5. 7,22,47,82,127,...7,22,47,82,127,...
6. 3,10,21,36,55,...3,10,21,36,55,...
7. 3,7,13,21,31,...3,7,13,21,31,...
8. 3,9,17,27,39,...3,9,17,27,39,...

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