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**Topic 8 Structures and Patterns II**

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**First degree: Tn = an+b e.g. 4,7,10,…..**

3 Tn d1 T1 4 3 T2 7 T3 10 T4 13 T5 16 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,…..

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**a=3 a+b = 4 ∴ b = 1 ∴ Tn = 3n + 1 Tn d1 T1 4 3 T2 7 T3 10 T4 13 T5 16**

Tn d1 T1 4 3 T2 7 T3 10 T4 13 T5 16 Tn=an+b d1 T1 a+b a T2 2a+b T3 3a+b T4 4a+b T5 5a+b a=3 a+b = 4 ∴ b = 1 ∴ Tn = 3n + 1

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**Second degree: Tn = an2+bn+c e.g. 6,13,26,45,70,…..**

7 T2 13 T3 26 19 T4 45 25 T5 70 Second degree: Tn = an2+bn+c e.g. 6,13,26,45,70,…..

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

Tn d1 T1 1a+1b+c 3a+b T2 4a+2b+c 2a 5a+b T3 9a+3b+c 7a+b T4 16a+4b+c 9a+b T5 25a+5b+c d2 First, let us examine the general case: Tn = an2 +bn+c

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**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 7a+b 16a+4b+c 9a+b 25a+5b+c 6 7 13 26 19 45 25 70 2a=6 a=3 3a+b = 7 ∴ b = -2 a+b+c = 6 ∴ c = 5

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**Exercises Find Tn and find T10**

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,….

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**If the values in this column are equal, then Tn = an3 + bn2 + cn + d**

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

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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+c 12a+2b 6a 19a+5b+c 18a+2b 37a+7b+c 6a 24a+2b 61a+9b+c

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**Exercises Find Tn and find T10**

1. 9,11,19,39,77,… ,-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,….

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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 25th term

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Sum of a sequence ….. to 25 terms 4 = = = = = 50 2a = 3 a = 1.5 3a + b = 7 4.5 + b = 7 b = 2.5 a + b + c = 4 c = 4 c = 0 Sn = 1.5n2+2.5n 7 3 10 3 13 3 16

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**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,…

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

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

25 of these

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

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

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

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

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

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

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

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

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

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**n = no. of units in each side K = total no. of squares **

1 2 5 3 14 4 30 ?

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6a = 2 a = 1/3 12a + 2b = 5 b = ½ 7a+3b+c = 4 c = 1/6 a+b+c+d = 1 d = 0 4 9 16 5 7 2

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**n P 1 2 4 3 10 20 Find the general rule n=1 n=2 n=3 n=4**

1 2 4 3 10 20 n=1 n=2 n=3 n=4 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

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n P 1 2 4 3 10 20 5 n=4 35 20 + 5 + 4 + 3 + 2 + 1

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**How many squares in the 20th figure?**

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**How many squares in the 20th figure?**

25 1 5 13

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**Tn = 2n2 -2n + 1 1 5 4 13 8 4 25 12 2nd differences are common**

∴ Tn = an2 + bn + c 1 5 13 25 4 8 12 a + b + c = 1 c = 1 2a = 4 a = 2 4 3a + b = 4 b = -2 Tn = 2n2 -2n + 1

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**Sequences other than APs and GPs**

Model: Write down the first 6 terms of the sequence in which t1=5 and tn+1=tn+3. tn+1=tn+3 t2 = t = 5+3 = 8 t3 = t = 8+3 = 11 t4 = t = 11+3 = 14 t5 = t = 14+3 = 17 t6 = t = 17+3 = 20 First 6 terms are 5, 8, 11, 14, 17, 20

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**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?

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

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**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)

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Fn r = Fn Fn-1 1 2 3 1.5 5 1.666… 8 1.6 13 1.625 21 34 1.619… 55 1.617… 89 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 surd

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** has some interesting properties. **

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

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**Do these numbers look familiar ?**

… and these also form a recursive function 1 1

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**Fibonacci Numbers in nature**

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

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Exercises NewQ Exercise 7.1 Page 177

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**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

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**Model: Prove that the sum of the first n squares, Sn= 12 + 22 + 32 +…**

Model: Prove that the sum of the first n squares, Sn= ….+n2 is given by Sn = Proof S1 = 12 = 1 Also S1 = = ∴ true for n=1 Assume true for n=k i.e. Sk= Now Sk+1 = Sk + (k+1)2 = + (k+1)2

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**Model: Prove that the sum of the first n squares, Sn= 12 + 22 + 32 +…**

Model: Prove that the sum of the first n squares, Sn= ….+n2 is given by Sn = = + (k+1)2 = + k2 + 2k + 1 Now 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

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**Exercise 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 = ….+n3 is given by 3. the sum of the first n fourth powers, Sn = …+n4 is given by

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Exercises NewQ Exercise Page 181

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**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 Assume true for n=k i.e. k(k+1)(k+2) = 6a (for some integer “a”) When n= k+1 Tk+1 = (k+1)(k+1+1)(k+1+2) = (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

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Exercises NewQ Exercise Page 181

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**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?

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Exercises NewQ Exercise 7.3 Page 187

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**Think of a two digit 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

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**Think of a two digit 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

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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!

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**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

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**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

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**THE ANSWER IS 22 Write down a three digit number**

452 45 42 52 54 25 24 242 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 = 11 Divide the previous total by this number Prove this THE ANSWER IS 22

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3 8 11 19 30 49 79 128 207 335 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 Prove this The answer is 11 x t7

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TERM a1 a2 a1 + a2 a1 + 2a2 a1 + 3a2 a1 + 5a2 a1 + 8a2 a1 + 13a2 a1 + 21a2 a1 + 34a2 55a1 + 88a2 = 11(5a1 + 8a2) = 11 x t7

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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 1089 Prove this x y 1089

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