3QuadraticSequences

114936254

743

Inductive Reasoning

1 2 3 4 5 6 … n … 20

0 3 10 21 36 55 … ? … ?

Quadratic Sequences

Terms: 4, 25, 36, 49, 64 …

3, 24, 35, 48, 63 …6, 12, 20, 30, 42 …

Factorable Quadratic Sequences

x 1 2 3 4 5 6 … x ... 20

y 6 12 20 30 42 56 … ...

This sequence is no longer linear. It is not constant till the 2nd level. It is quadratic as can seen in the graph of the points.

6 8 10 12 142 2 2 2

Gap

GapNo common Gap !

As you can see, this graph is part of a parabola or quadratic equation.

If we examine some characteristics of quadratic equations, we will be able to see how these equations can be converted into two linear sequences for which it is very easy to find formulas.

Solving Quadratic Equations

2 8 7 0x x

( 7)( 1) 0x x

Set = to 0, then factor.

The two factors are in y = mx + b form.

If we factor the terms into linear sequences, then each factor can be easily converted into an algebraic expression in the

form or mx + b.

Wait !

1*6

2*31*122*63*4 4*5

2*10

1*20

x 1 2 3 4 5 6 … x ... 20

y 6 12 20 30 42 56 … ...

Only certain combination will create two linear sequences. Our difficulty is determining which factors will create the sequences.

6 8 10 12 142 2 2 2

Break terms into factors

Gap

Gap

1*6

2*31*122*63*4 4*5

2*10

1*205*6 6*7

x 1 2 3 4 5 6 … x ... 20

y 6 12 20 30 42 56 … ...

Let’s see if the pattern continues.

6 8 10 12 142 2 2 2

Do you see the pattern ?

7*8

It does !

Gap

Gap

2*33*4 4*5

5*6 6*7

x 1 2 3 4 5 6 … x ... 20

y 6 12 20 30 42 56 … ...

For the green sequence, the gap between terms is 1. Therefore the slope is 1.

6 8 10 12 142 2 2 2

Now find the formula for the green sequence.

7*8

2 = (1)1 + b 1 = b (x + 1)

(x + 1)

Gap

Gap

2*33*4 4*5

5*6 6*7

x 1 2 3 4 5 6 … x ... 20

y 6 12 20 30 42 56 … ...

For the red sequence, the gap between terms is also 1. Therefore the slope is 1.

6 8 10 12 142 2 2 2

Now find the formula for the red sequence.

7*8

3 = (1)1 + b 2 = b (x + 2)

(x + 1)(x + 2)

Gap

Gap

2*33*4 4*5

5*6 6*7

x 1 2 3 4 5 6 … x ... 20

y 6 12 20 30 42 56 … ...

Substitute 20 for x.

6 8 10 12 142 2 2 2

Now let’s find the 20th term.

7*8(x + 1)(x + 2)

Gap

Gap

(x + 2)(x + 1)

(21)(22) = 462

Let’s try another problem. x 1 2 3 4 5 6 … x ... 20

y 0 7 16 27 40 55 … ...

7 9 11 13 152 2 2 2

Gap

Gap

Factor the terms

1*72*63*4 4*4

2*8

1*16

x 1 2 3 4 5 6 … x ... 20

y 0 7 16 27 40 55 … ...

7 9 11 13 152 2 2 2

Gap

Gap

3*9

1*27

Do you see the pattern?

SkipToomany

Test to see if it really works.

1*72*8

x 1 2 3 4 5 6 … x ... 20

y 0 7 16 27 40 55 … ...

7 9 11 13 152 2 2 2

Gap

Gap

3*9

It really does work.

SkipToomany

4*10 5*11

Now find the formula for the green sequence.

1*72*8

x 1 2 3 4 5 6 … x ... 20

y 0 7 16 27 40 72 … ...

7 9 11 13 152 2 2 2

Gap

Gap

3*9SkipToomany

4*10 5*11

For the green sequence, the gap between terms is 1. Therefore the slope is 1.1 = (1)2 + b -1 = b (x-1)

(x-1)

Now find the formula for the red sequence.

1*72*8

x 1 2 3 4 5 6 … x ... 20

y 0 7 16 27 40 72 … ...

7 9 11 13 152 2 2 2

Gap

Gap

3*9SkipToomany

4*10 5*11

For the red sequence, the gap between terms is 1. Therefore the slope is 1.7 = (1)2 + b 5 = b (x + 5)

(x-1)(x + 5)

Let’s find the 20th term

1*72*8

x 1 2 3 4 5 6 … x ... 20

y 0 7 16 27 40 72 … ...

7 9 11 13 152 2 2 2

Gap

Gap

3*9SkipToomany

4*10 5*11

(x-1)(x + 5)

475

(x-1)(x + 5)

(19)(25) =

Elimination Strategies

All linear sequence either constantly increase or constantly decrease.

Therefore if the sequence doesn’t continually increase or decrease, the factors can be rejected.

Knowing this will increase the speed of finding the correct set of factors..

Revisit first problem.

1*6

2*31*122*63*4

x 1 2 3 4 5 6 … x ... 20

y 6 12 20 30 42 56 … ...

6 8 10 12 142 2 2 2

Gap

Gap1* 12 can be immediately rejected because the green factors do not change.2* 6 can be immediately rejected because either way the green factors do not change or the red factors do not change .

2*31*122*63*4

x 1 2 3 4 5 6 … x ... 20

y 6 12 20 30 42 56 … ...

6 8 10 12 142 2 2 2

Gap

Gap

4*5 5*6 6*7

1*6

Doesn’t work because the values do not

increase in the red factors.

2*31*122*63*4

x 1 2 3 4 5 6 … x ... 20

y 6 12 20 30 42 56 … ...

6 8 10 12 142 2 2 2

Gap

Gap

Applying the pattern of increasing each factor by 1 each time, we can predict the next factor terms.

1*6

1*6

2*31*122*63*4

x 1 2 3 4 5 6 … x ... 20

y 6 12 20 30 42 56 … ...

6 8 10 12 142 2 2 2

Gap

Gap

4*5 5*6 6*7

Note that now we did not need to try as many factor combinations.

Now it is easy to compute the linear formula for each factor sequence.

By using the concept that the linear factors need to increase or decrease every time, the number of trials to find the correct sequence of factors is greatly reduced.

Revisit second problem.

1*72*63*4 4*4

2*8

1*16

x 1 2 3 4 5 6 … x ... 20

y 0 7 16 27 40 72 … ...

7 9 11 13 152 2 2 2

Gap

Gap

SkipToomany

1* 16 can be immediately rejected because the green factors do not change.4* 4 can be immediately rejected because either way the green factors do not change or the red factors do not change properly.

Revisit second problem.

1*72*63*4 4*4

2*8

1*16

x 1 2 3 4 5 6 … x ... 20

y 0 7 16 27 40 55 … ...

7 9 11 13 152 2 2 2

Gap

Gap

3*9SkipToomany

The pattern is quickly discovered. Therefore the next factors can be predicted and tested.

4*10 5*11

Revisit second problem.

1*72*63*4 4*4

2*8

1*16

x 1 2 3 4 5 6 … x ... 20

y 0 7 16 27 40 55 … ...

7 9 11 13 152 2 2 2

Gap

Gap

3*9SkipToomany

The pattern works. Now we could compute the linear formulas for each factor sequence. But we will not at this time.

4*10 5*11

Now we need to practice.

Let’s begin.

Practice 1a x 1 2 3 4 5 6 … x ... 20

y 0 5 12 21 32 45 … ...

5 7 9 11 132 2 2 2

Gap

Gap

Factor each term.

Practice 1b

1*52*63*4 3*4

2*6

1*12

x 1 2 3 4 5 6 … x ... 20

y 0 5 12 21 32 45 … ...

2 2 2 2Gap

Gap

SkipToomany

1* 12 can be immediately rejected because the green factors do not change.3* 4 can be immediately rejected because either way the green factors do not change or the red factors do not change .

5 7 9 11 13

Practice 1c

1*52*63*4 3*4

2*6

1*12

x 1 2 3 4 5 6 … x ... 20

y 0 5 12 21 32 45 … ...

2 2 2 2Gap

Gap

3*7SkipToomany

4*8 5*9

The pattern is quickly discovered. Therefore the next factors can be predicted and tested.

It really does works.

5 7 9 11 13

Practice 1d

1*52*6

x 1 2 3 4 5 6 … x ... 20

y 0 5 12 21 32 45 … ...

2 2 2 2Gap

Gap

3*7SkipToomany

4*8 5*9

For the green sequence, the gap between terms is 1. Therefore the slope is 1.1 = (1)2 + b -1 = b (x-1)

(x-1)

5 7 9 11 13

Practice 1e

1*52*6

x 1 2 3 4 5 6 … x ... 20

y 0 5 12 21 32 45 … ...

2 2 2 2Gap

Gap

3*7SkipToomany

4*8 5*9

For the red sequence, the gap between terms is 1. Therefore the slope is 1.

5 = (1)2 + b 3 = b (x+ 3)

(x-1)(x+ 3)

5 7 9 11 13

Practice 1f

1*52*6

x 1 2 3 4 5 6 … x ... 20

y 0 5 12 21 32 45 … ...

2 2 2 2Gap

Gap

3*7SkipToomany

4*8 5*9

(x-1)

(x+ 3) Let’s find the 20th term(x-1)(19) (23) = 437

437

(x+ 3)

5 7 9 11 13

Practice 2a x 1 2 3 4 5 6 … x ... 20

y 18 28 40 54 70 88 … ...

10 12 14 16 182 2 2 2

Gap

Gap

Factor each term.

1*28

Practice 2b

1*182*144*7

5*8

x 1 2 3 4 5 6 … x ... 20

y 18 28 40 54 70 88 … ...

2 2 2 2Gap

Gap

6*9 7*10 8*112*93*6

1* 28 can be immediately rejected because the green factors do not change.2* 14 can be immediately rejected because either way the green factors do not change or the red factors do not change .

10 12 14 16 18

1*28

4*73*6

Practice 2c

1*182*14

5*8

x 1 2 3 4 5 6 … x ... 20

y 18 28 40 54 70 88 … ...

2 2 2 2Gap

Gap

6*9 7*10 8*112*9

The pattern is quickly discovered. Therefore the next factors can be predicted and tested.

It really does works.

10 12 14 16 18

Practice 2d

4*75*8

x 1 2 3 4 5 6 … x ... 20

y 18 28 40 54 70 88 … ...

2 2 2 2Gap

Gap

6*9 7*10 8*113*6

For the green sequence, the gap between terms is 1. Therefore the slope is 1.3 = (1)1 + b 2 = b (x + 2)

(x + 2)

10 12 14 16 18

Practice 2e

4*75*8

x 1 2 3 4 5 6 … x ... 20

y 18 28 40 54 70 88 … ...

2 2 2 2Gap

Gap

6*9 7*10 8*113*6

For the red sequence, the gap between terms is 1. Therefore the slope is 1.6 = (1)1 + b 5 = b (x + 5)

(x + 2)(x + 5)

10 12 14 16 18

Practice 2f

4*75*8

x 1 2 3 4 5 6 … x ... 20

y 18 28 40 54 70 88 … ...

2 2 2 2Gap

Gap

6*9 7*10 8*113*6

(x+2)(x + 5)

Let’s find the 20th term(x + 5)(x+2)(22) (25) = 550

550

10 12 14 16 18

Practice 3a x 1 2 3 4 5 6 … x ... 20

y 20 30 42 56 72 90 … ...

2 2 2 2Gap

Gap

Factor each term.

10 12 14 16 18

Practice 3b

1*302*15

x 1 2 3 4 5 6 … x ... 20

y 20 30 42 56 72 90 … ...

2 2 2 2Gap

Gap

1*202*10

4*5

1* 30 can be immediately rejected because the green factors do not change.2* 15 can be immediately rejected because either way the green factors do not change or the red factors do not change .

10 12 14 16 18

3*10

Practice 3c

1*72*15

x 1 2 3 4 5 6 … x ... 20

y 20 30 42 56 72 90 … ...

2 2 2 2Gap

Gap

1*202*10

4*5

3* 10 can be immediately rejected because either way the green factors do not change or the red factors do not change .

5*12

10 12 14 16 18

5*64*5

Practice 3d

1*72*153*10 6*7

x 1 2 3 4 5 6 … x ... 20

y 20 30 42 56 72 90 … ...

2 2 2 2Gap

Gap

7*8 8*9 9*10

1*202*10

The pattern is quickly discovered. Therefore the next factors can be predicted and tested.

It really does works.

10 12 14 16 18

Practice 3e

6*7

x 1 2 3 4 5 6 … x ... 20

y 20 30 42 56 72 90 … ...

2 2 2 2Gap

Gap

7*8 8*9 9*104*5 5*6

For the green sequence, the gap between terms is 1. Therefore the slope is 1.4 = (1)1 + b 3 = b (x + 3)

(x + 3)

10 12 14 16 18

Practice 3f

6*7

x 1 2 3 4 5 6 … x ... 20

y 20 30 42 56 72 90 … ...

2 2 2 2Gap

Gap

7*8 8*9 9*104*5 5*6

(x + 3)

For the red sequence, the gap between terms is 1. Therefore the slope is 1.5 = (1)1 + b 4 = b (x + 4)

(x + 4)

10 12 14 16 18

Practice 3g

6*7

x 1 2 3 4 5 6 … x ... 20

y 20 30 42 56 72 90 … ...

2 2 2 2Gap

Gap

7*8 8*9 9*104*5 5*6

(x + 3)

Let’s find the 20th term

(x + 4)

(x + 3) (x + 4)

(23) (24) = 552

552

10 12 14 16 18

Practice 4a x 1 2 3 4 5 6 … x ... 20

y 0 9 20 33 48 65 … ...

9 11 13 15 172 2 2 2

Gap

Gap

SkipToomany

Factor each term.

Practice 4b

1*93*3

4*52*10

1*20

x 1 2 3 4 5 6 … x ... 20

y 0 9 20 33 48 65 … ...

2 2 2 2Gap

Gap

SkipToomany

1* 20 can be immediately rejected because the green factors do not change.

9 11 13 15 17

Let’s try the pattern of 1*9 and 2*10.

Practice 4c

1*93*3

4*52*10

1*20

x 1 2 3 4 5 6 … x ... 20

y 0 9 20 33 48 65 … ...

2 2 2 2Gap

Gap

3*11SkipToomany

4*12 5*13

First try is a charm.

It really works.9 11 13 15 17

Practice 4d x 1 2 3 4 5 6 … x ... 20

y 0 9 20 33 48 65 … ...

2 2 2 2Gap

Gap

SkipToomany

For the green sequence, the gap between terms is 1. Therefore the slope is 1.1 = (1)2 + b -1 = b (x - 1)

9 11 13 15 17

1*92*10 3*11 4*12 5*13

(x - 1)

Practice 4e

1*92*10

x 1 2 3 4 5 6 … x ... 20

y 0 9 20 33 48 65 … ...

2 2 2 2Gap

Gap

SkipToomany

For the red sequence, the gap between terms is 1. Therefore the slope is 1.9 = (1)2 + b 7 = b (x + 7)

9 11 13 15 17

3*11 4*12 5*13

(x - 1)(x + 7)

Practice 4f

1*92*10

x 1 2 3 4 5 6 … x ... 20

y 0 9 20 33 48 65 … ...

2 2 2 2Gap

Gap

SkipToomany

Let’s find the 20th term

9 11 13 15 17

3*11 4*12 5*13

(x - 1)(x + 7)

(x + 7)(x - 1)(19) (27) = 513

513

Summary When the gap between terms of sequences are constant on the second level, the sequence is quadratic or second degree.

The rule for each term is created by the factors of each term which create their own linear sequence.

The quadratic sequence is treated as the product of 2 linear sequences.

Time can be saved by assuming a pattern exists in the first two columns of factors, then seeing if the terms can correctly predict the following terms.

Also, time can be saved by rejecting terms that do not increase or decrease.

These problems require a lot of practice.

C’est fini.Good day and good luck.

A Senior Citizen Production

That’s all folks.

Good day and good luck.

C’est fini.