Topic 8Structures and Patterns II

Problem Given the first few terms of a sequence, to find an expression for Tn, the general term:

First degree: Tn = an+b

e.g. 4,7,10,…..

3 

Tn d1

T1 4

3

T2 7

3

T3 10

3

T4 13

3

T5 16 

3 

Tn d1

T1 4

3

T2 7

3

T3 10

3

T4 13

3

T5 16 

Tn=an+b

d1

T1 a+b

a

T2 2a+b

a

T3 3a+b

a

T4 4a+b

a

T5 5a+b 

a=3

a+b = 4 ∴ b = 1

∴ Tn = 3n + 1

Second degree:

Tn = an2+bn+c

e.g.

6,13,26,45,70,…..

Tn d1

T1 6

 

7

T2 13 6

13

T3 26 6

19

T4 45 6

25

T5 70 

First, let us examine the general case: Tn = an2 +bn+c

 

Tn d1

T11a+1b+

c

 

3a+b

T24a+2b+

c

2a

5a+b

T39a+3b+

c

2a

7a+b

T416a+4b

+c

2a

9a+b

T525a+5b

+c 

d2

3 

2a=6 a=3

3a+b = 7 ∴ b = -2

a+b+c = 6 ∴ c = 5

1a+1b+c

 

3a+b

4a+2b+c2a

5a+b

9a+3b+c2a

7a+b

16a+4b+c

2a

9a+b

25a+5b+c

 

6

 

7

13 6

13

26 6

19

45 6

25

70 

1.1,5,9,….2.6,11,16,….3.2,2,4,8,14,….4.1,15,35,61,93,….5.2,7,16,29,46,….

ExercisesExercises Find TFind Tn n and find T and find T1010

 

If neither of the difference columns, d1 or d2 have equal values, proceed to the third difference column (d3)

If the values in this column are equal, then

Tn = an3 + bn2 + cn + d

 

Exercise: Find the first 5 general terms of a third degree sequence and show d1, d2 and d3

Tn = an3 + bn2 + cn + d

T1 = a+b+c+d

T2 = 8a+4b+2c+d

T3 = 27a+9b+3c+d

T4 = 64a+16b+4c+d

T5 = 125a+25b+5c+d

7a+3b+c19a+5b+c37a+7b+c61a+9b+c

12a+2b

18a+2b

24a+2b

6a

6a

Exercises Find Tn and find T10

1. 9,11,19,39,77,….2. -8,-3,18,67,156,….3. 20,45,112,245,468,….4. 1058,1410,2126,3318,5134,7758,11410….5. -2,45,178,469,1014,1933,3370,….6. 22,130,554,1630,3814,7682,13930,….7. -687,-254,484,1634,3351,5838,9346,….

Sum of a sequence

Given a sequence, find the sum of the terms in that sequence.

e.g. 4+7+10+13+….. to 25 terms

Note: We are not looking for the 25th term

Sum of a sequence

4+7+10+13+….. to 25 terms

4 = 4

4+7 = 11

4+7+10 = 21

4+7+10+13 = 34

4+7+10+13+16 = 50

7

16

3

13

103

3

2a = 3 a = 1.5

3a + b = 74.5 + b = 7 b = 2.5

a + b + c = 41.5+2.5+ c = 4 c = 0

Sn = 1.5n2+2.5n

Exercises Find the sum of the first 20 terms of the sequences below:

1. 2,7,12,17,…2. 1,4,9,16,….3. 6,12,22,36,54,…4. 2,4,14,38,82,152,…

How many squares (of any size) can be seen in the figure below?

How many squares (of any size) can be seen in the figure below?

25 of these

How many squares (of any size) can be seen in the figure below?

How many squares (of any size) can be seen in the figure below?

How many squares (of any size) can be seen in the figure below?

How many squares (of any size) can be seen in the figure below?

How many squares (of any size) can be seen in the figure below?

How many squares (of any size) can be seen in the figure below?

How many squares (of any size) can be seen in the figure below?

How many squares (of any size) can be seen in the figure below?

etc

16 of these

and so on….

Is there a bette

r way?

n = no. of units in each sideK = total no. of squares

nn KK

11 11

22 55

33 1414

44 3030

55 ??

1

5

14

30

44

99

1166

55

7722

6a = 2

a = 1/3

12a + 2b = 5

b = ½

7a+3b+c = 4

c = 1/6

a+b+c+d = 1

d = 0

nnnTn 612

213

31

55

555 612

213

31

5

T

n=1

n=2

n=3

n=4

nn PP

00 00

11 11

22 44

33 1010

44 2020In the figures above,

n = the no. of units on each side

P = total no. of “point-up” triangles of all sizes

Find the general rule

n=4

20 +

nn PP

00 00

11 11

22 44

33 1010

44 2020

55 5 + 4 + 3+ 2 + 1 3535

How many squares in the 20th figure?

How many squares in the 20th figure?

1 5 13

25

1

5

13

25

4

8

12

4

4

2nd differences are common

∴ Tn = an2 + bn + c

2a = 4

a = 2

3a + b = 4

b = -2

a + b + c = 1

c = 1

Tn = 2n2 -2n + 1

Sequences other than APs and GPsModel: Write down the first 6 terms of the

sequence in which t1=5 and tn+1=tn+3.

tn+1=tn+3t2 = t1+3 = 5+3 = 8t3 = t2+3 = 8+3 = 11t4 = t3+3 = 11+3 = 14t5 = t4+3 = 14+3 = 17t6 = t5+3 = 17+3 = 20

First 6 terms are 5, 8, 11, 14, 17, 20

The Fibonacci Sequence

The Fibonacci sequence was derived by Leonardo of Pisa who used the name Fibonacci for his published writings.

The question that Leonardo posed that led to the development of the Fibonacci sequence was this one:

How many pairs of rabbits can be produced from a single pair in one year if it is assumed that every month each pair begets a new pair which from the second month becomes productive?

A

A

B

E

C

A

A

A

B

C

B

D

after 5 months

after 6months

Fibonacci Sequence

f1=1 f2=1 fn+2=fn+1+fn

First terms are 1, 1, 2, 3, 5, 8, 13, 21, ……

Read top of P175

Investigation P175 (New Q)

FFnn r = Fr = Fnn F Fn-n-

11

11

11 11

22 22

33 1.51.5

55 1.666…1.666…

88 1.61.6

1313 1.6251.625

2121 1.615..1.615..

3434 1.619…1.619…

5555 1.617…1.617…

8989 1.618…1.618…

Use the statistical graphing capability of your calculator to produce a scatter graph of r against n

The approximate value of the limit of r is known as (phi) which is written as the surd2

15

has some interesting properties.e.g. Consider what happens when you raise

to increasing powers:

2

1155

2

15

2

753

2

15

2

753

2

15

2

452

2

15

2

452

2

15

2

35

2

15

2

35

2

15

2

15

5

5

4

4

3

3

2

2

Note:

+ 2 = 3

2 + 3 = 4

Note also

1 + = 2

This means that

1, , 2, 3, 4 form a Fibonacci sequence

2

1155

2

15

2

753

2

15

2

753

2

15

2

452

2

15

2

452

2

15

2

35

2

15

2

35

2

15

2

15

5

5

4

4

3

3

2

2

1

1

Do these numbers look familiar ?

… and these also form a recursive function

Fibonacci Numbers in nature

The scales on a pine cone (and a pineappple) are arranged in spirals of Fibonacci numbers

NewQ

Exercise 7.1

Page 177

Proof by InductionSteps:• Prove true for n=1• State the proposition for n=k• Assume the truth of the proposition for n=k and

show that it is true for n=k+1• If true for n=1, then it must be true for n=2• If true for n=2, then it must be true for n=3 • etc

Model: Prove that the sum of the first n squares, Sn= 12 + 22 + 32 +….+n2 is given by

Sn =

ProofS1 = 12 = 1

Also S1 =

= =

∴ true for n=1

Assume true for n=ki.e. Sk=

Now Sk+1 = Sk + (k+1)2

= + (k+1)2

nnn 612

213

31

1

111

61

21

31

612

213

31

kkk 612

213

31

kkk 612

213

31

Model: Prove that the sum of the first n squares, Sn= 12 + 22 + 32 +….+n2 is given by

Sn =

= + (k+1)2 = + k2 + 2k + 1

Now

nnn 612

213

31

kkk 612

213

31

kkk 612

213

31

)61392(

)612632(

12

2361

22361

6122

213

31

kkk

kkkkk

kkkkk

)61392(

)13632662(

)1()12()133(

)1()1()1(

2361

22361

612

2123

31

612

213

31

kkk

kkkkkk

kkkkkk

kkk

i.e. Rule is true for Sk+1

∴ If true for n=1 then true for n=2

If true for n=2 then true for n=3

etc Sk+1

Use the method of proof by induction to prove:

1. the sum of the first n terms of a GP with first term, a, and common ratio, r,

is given by

2. the sum of the first n cubes, Sn = 13+23+33+….+n3 is given by

r

raS

n

n

1

)1(

2413

214

41 nnnSn

3. the sum of the first n fourth powers, Sn = 14+24+34+…+n4 is

given by )133)(12)(1( 230 nnnnS n

n

NewQ

Exercise 7.2 1-3

Page 181

Model : Show that n(n+1)(n+2) is divisible by 6When n=1, n(n+1)(n+2) = 1x2x3 = 6

∴ true for n=1

When n= k+1Tk+1 = (k+1)(k+1+1)(k+1+2)

Assume true for n=ki.e. k(k+1)(k+2) = 6a (for some integer “a”)

= (k+1)(k+2)(k+3) = k(k+1)(k+2) + 3(k+1)(k+2)

= 6a + 3(k+1)(k+2) which must be a multiple of 6 since (k+1)(k+2) must be even

= 6a + 6b for some integer “b” = 6(a+b)

∴ true for n=k+1 etc

NewQ

Exercise 7.2 4-7

Page 181

Hailstone sequences (p183)

Sequences such as 5, 16, 8, 4, 2, 1, 4, 2, 1, … and 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, …are known as hailstone sequences because they bounce around

before coming to rest.Hailstone sequences are generated as follows:• Start with any positive integer n (Everybody choose one)

• If n is even, divide by 2 to get the next term• If n is odd, multiply by 3 and add 1 to get the next term• Repeat this process with successive terms (Everybody try this)

One of mathematics’ unsolved problems is to prove that every starting value will generate a sequence that eventually settles to 4, 2, 1, 4, 2, 1, …Could there possibly be a sequence that doesn’t settle down to this cycle?

NewQ

Exercise 7.3

Page 187

Think of a two digit number.Add together these 2 digits and subtract this sum from your original number.

When you have the final number look it up on the chart below:

I will now tell you the symbol associated with your number

Think of a two digit number.Add together these 2 digits and subtract this sum from your original number.

When you have the final number look it up on the chart below:

I will now tell you the symbol associated with your number

Methods of Proof• Proof by obviousness The proof is so clear that it need not

be mentioned

• Proof by general agreement All those in favour

• Proof by imagination We’ll pretend that it’s true…

• Proof by necessity It had better be true or the entire structure of mathematics would crumble to the ground.

• Proof by plausibility It sounds good, so it must be true.

• Proof by intimidation Don’t be stupid. Of course it’s true!

• Proof by lack of sufficient Because of the time constraint, I’ll time leave the proof to you.

• Proof by postponement Because the proof of this is so long,it is given in the appendix.

• Proof by accident Hey! What have we got here?

• Proof by insignificance Who cares anyway?

• Proof by profanity (Example censored)

• Proof by definition We define it to be true

• Proof by lost reference I know I saw it somewhere

• Proof by calculus This proof requires calculus, so we’ll skip it

• Proof by lack of interest Does anyone really want to see this?

• Proof by illegibility

• Proof by divine word And the Lord said, “Let it be true” and it was true

• Proof by intuition I just have this gut feeling

• Write down a three digit number• Write down as many 2 digit combinations of

these 3 digits as possible and add these numbers• Add the digits of the original number 4+5+2 = 11Divide the previous total by this number

452

45

42

52

54

25

24

242

THE ANSWER IS 22

Prove this

• Write down any 10 numbers that have a Fibonacci type sequence i.e. tn + tn+1 = tn+2

• Add these numbers

• Let me see your numbers and I will quickly tell you what they add up to

3

8

11

19

30

49

79

128

207

335

The answer is 11 x t7

Prove this

TERM1 a1

2 a2

3 a1 + a2

4 a1 + 2a2

5 2a1 + 3a2

6 3a1 + 5a2

7 5a1 + 8a2

8 8a1 + 13a2

9 13a1 + 21a2

10 21a1 + 34a2

55a1 + 88a2 = 11(5a1 + 8a2) = 11 x t7

Think of a 3 digit numberReverse the digits and take the smaller from the larger

Call this number xReverse the digits of x to give you another 3 digit number.

Call this number y.Add x and y

Your answer is 1089

952- 259 693 x+ 396 y 1089

Prove this