11 49 36 25 4 7 4 3 Inductive Reasoning 1 2 3 4 5 6 … n … 20 0 3 10 21 36 55 … ? … ?
3QuadraticSequences
Jun 27, 2015
Algebraic Way of finding the nth term of the sequenc
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
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.