Download presentation

Presentation is loading. Please wait.

Published byEthan Stevenson Modified over 3 years ago

1
Topic 8 Structures and Patterns II

2
Problem Given the first few terms of a sequence, to find an expression for T n, the general term: First degree: T n = an+b e.g. 4,7,10,….. 3 TnTn d1d1 T1T1 4 3 T2T2 7 3 T3T T4T T5T5 16

3
3 TnTn d1d1 T1T1 4 3 T2T2 7 3 T3T T4T T5T5 16 T n =an+b d1d1 T1T1 a+b a T2T2 2a+b a T3T3 3a+b a T4T4 4a+b a T5T5 5a+b a=3 a+b = 4 b = 1 T n = 3n + 1

4
Second degree: T n = an 2 +bn+c e.g. 6,13,26,45,70,….. TnTn d1d1 T1T1 6 7 T2T2 136 T3T T4T T5T5 70

5
First, let us examine the general case: T n = an 2 +bn+c TnTn d 1 T1T1 1a+1b+c 3a+b T2T2 4a+2b+c 2a 5a+b T3T3 9a+3b+c 2a 7a+b T4T4 16a+4b+c 2a 9a+b T5T5 25a+5b+c d2d2

6
3 2a=6 a=3 3a+b = 7 b = -2 a+b+c = 6 c = 5 1a+1b+c 3a+b 4a+2b+c 2a 5a+b 9a+3b+c 2a 7a+b 16a+4b+c 2a 9a+b 25a+5b+c

7
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,…. Exercises Find T n and find T 10

8
If neither of the difference columns, d 1 or d 2 have equal values, proceed to the third difference column (d 3 ) If the values in this column are equal, then T n = an 3 + bn 2 + cn + d

9
Exercise: Find the first 5 general terms of a third degree sequence and show d 1, d 2 and d 3 T n = an 3 + bn 2 + cn + d T 1 = a+b+c+d T 2 = 8a+4b+2c+d T 3 = 27a+9b+3c+d T 4 = 64a+16b+4c+d T 5 = 125a+25b+5c+d 7a+3b+c 19a+5b+c 37a+7b+c 61a+9b+c 12a+2b 18a+2b 24a+2b 6a

10
Exercises Find T n and find T ,11,19,39,77,… ,-3,18,67,156,… ,45,112,245,468,… ,1410,2126,3318,5134,7758,11410… ,45,178,469,1014,1933,3370,… ,130,554,1630,3814,7682,13930,… ,-254,484,1634,3351,5838,9346,….

11

12
Sum of a sequence Given a sequence, find the sum of the terms in that sequence. e.g ….. to 25 terms Note: We are not looking for the 25 th term

13
Sum of a sequence ….. to 25 terms 4= 4 4+7= = = = a = 3 a = 1.5 3a + b = b = 7 b = 2.5 a + b + c = c = 4 c = 0 S n = 1.5n n

14

15
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,…

16
How many squares (of any size) can be seen in the figure below?

17
25 of these

18
How many squares (of any size) can be seen in the figure below?

19

20

21

22

23

24

25
etc 16 of these and so on…. Is there a better way?

26
n = no. of units in each side K = total no. of squaresnK ?

27
a = 2 a = 1/3 12a + 2b = 5 b = ½ 7a+3b+c = 4 c = 1/6 a+b+c+d = 1 d = 0

28
n=1 n=2 n=3 n=4 nP In 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

29
n= nP

30
How many squares in the 20 th figure?

31

32
nd differences are common T n = an 2 + bn + c 2a = 4 a = 2 3a + b = 4 b = -2 a + b + c = 1 c = 1 T n = 2n 2 -2n + 1

33
Sequences other than APs and GPs Model: Write down the first 6 terms of the sequence in which t 1 =5 and t n+1 =t n +3. t n+1 =t n +3 t 2 = t 1 +3 = 5+3= 8 t 3 = t 2 +3 = 8+3= 11 t 4 = t 3 +3 = 11+3= 14 t 5 = t 4 +3 = 14+3= 17 t 6 = t 5 +3 = 17+3= 20 First 6 terms are 5, 8, 11, 14, 17, 20

34
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?

35
A A B E C A A A B C B D after 5 months after 6months

36
Fibonacci Sequence f 1 =1 f 2 =1 f n+2 =f n+1 +f n First terms are 1, 1, 2, 3, 5, 8, 13, 21, …… Read top of P175 Investigation P175 (New Q)

37

38
FnFnFnFn r = F n F n … … … … 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 surd

39
has some interesting properties. e.g. Consider what happens when you raise to increasing powers: Note: + 2 = = 4 Note also 1 + = 2 This means that 1,, 2, 3, 4 form a Fibonacci sequence

40
1 1 Do these numbers look familiar ? … and these also form a recursive function

41
Fibonacci Numbers in nature

42
The scales on a pine cone (and a pineappple) are arranged in spirals of Fibonacci numbers

43

44

45

46
NewQ Exercise 7.1 Page 177

47
Proof by Induction Steps: 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

48
Model: Prove that the sum of the first n squares, S n = ….+n 2 is given by S n = Proof S 1 = 1 2 = 1 Also S 1 = = = true for n=1 Assume true for n=k i.e. S k = Now S k+1 = S k + (k+1) 2 = + (k+1) 2

49
Model: Prove that the sum of the first n squares, S n = ….+n 2 is given by S n = = + (k+1) 2 = + k 2 + 2k + 1 Now i.e. Rule is true for S k+1 If true for n=1 then true for n=2 If true for n=2 then true for n=3 etc S k+1

50
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, S n = ….+n 3 is given by 3. the sum of the first n fourth powers, S n = …+n 4 is given by

51
NewQ Exercise Page 181

52
Model : Show that n(n+1)(n+2) is divisible by 6 When n=1, n(n+1)(n+2) = 1x2x3 = 6 true for n=1 When n= k+1 T k+1 = (k+1)(k+1+1)(k+1+2) Assume true for n=k i.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

53
NewQ Exercise Page 181

54
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 doesnt settle down to this cycle?

55
NewQ Exercise 7.3 Page 187

56
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

57
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

58
Methods of Proof Proof by obviousnessThe proof is so clear that it need not be mentioned Proof by general agreementAll those in favour Proof by imaginationWell pretend that its true… Proof by necessityIt had better be true or the entire structure of mathematics would crumble to the ground. Proof by plausibilityIt sounds good, so it must be true. Proof by intimidationDont be stupid. Of course its true!

59
Proof by lack of sufficientBecause of the time constraint, Ill timeleave the proof to you. Proof by postponementBecause the proof of this is so long, it is given in the appendix. Proof by accidentHey! What have we got here? Proof by insignificanceWho cares anyway? Proof by profanity(Example censored) Proof by definitionWe define it to be true Proof by lost referenceI know I saw it somewhere Proof by calculusThis proof requires calculus, so well skip it

60
Proof by lack of interestDoes anyone really want to see this? Proof by illegibility Proof by divine wordAnd the Lord said, Let it be true and it was true Proof by intuitionI just have this gut feeling

61
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 = 11 Divide the previous total by this number THE ANSWER IS 22 Prove this

62
Write down any 10 numbers that have a Fibonacci type sequence i.e. t n + t n+1 = t n+2 Add these numbers Let me see your numbers and I will quickly tell you what they add up to The answer is 11 x t 7 Prove this

63
TERM 1 a 1 2 a 2 3 a 1 + a 2 4 a 1 + 2a 2 5 2a 1 + 3a 2 6 3a 1 + 5a 2 7 5a 1 + 8a 2 8 8a a a a a a 2 55a a 2 = 11(5a 1 + 8a 2 ) = 11 x t 7

64
Think of a 3 digit number Reverse the digits and take the smaller from the larger Call this number x Reverse the digits of x to give you another 3 digit number. Call this number y. Add x and y Your answer is x y 1089 Prove this

65

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google