1. Chapter 11 Sec 6 Recursion and Special Sequences

2. 2 of 17 Algebra 2 Chapter 11 Sections 6 & 7 Fibonacci Sequence in Nature A shoot on a sneezewort plant must grow for two months before it is strong enough to put out another shoot. After that it puts out at least one shoot every month. Notice that the sequence 1, 1, 2, 3, 5, 8, 13,..has a pattern. Each term in the sequence is the sum of the two previous terms. This is called a Fibonacci Sequence. First terma11 Second terma21 Third terma3a1 + a21 + 1 = 2 Nth termanan – 2 + an – 1

3. 3 of 17 Algebra 2 Chapter 11 Sections 6 & 7 Recursive Formula The formula a n = a n – 2 + a n – 1 is an example of a recursive formula. This means that each term is formulated from one or more previous terms. To be able to use a recursive formula, you must be given the value(s) of the first term(s) so that you can start the sequence and then use the formula to generate the rest of the terms. The formula a n = a n – 2 + a n – 1 is an example of a recursive formula. This means that each term is formulated from one or more previous terms. To be able to use a recursive formula, you must be given the value(s) of the first term(s) so that you can start the sequence and then use the formula to generate the rest of the terms. Ex. Find the first five terms of the sequence in which a 1 = 4 and a n + 1 = 3a n – 2, n > 1. Ex. Find the first five terms of the sequence in which a 1 = 4 and a n + 1 = 3a n – 2, n > 1. a 1 + 1 = 3a 1 – 2 a 2 = 3(4) – 2 = 10 a 2 + 1 = 3(10) – 2 a 3 = 28 a 3 + 1 = 3(28) – 2 a 4 = 82 a 4 + 1 = 3(82) – 2 a 5 = 244 4, 10, 28, 82, 244

4. 4 of 17 Algebra 2 Chapter 11 Sections 6 & 7 Example 1 Find the first five terms of the sequence in which a 1 = 5 and a n + 1 = 2a n + 7, n > 1. a n + 1 = 2a n + 7 a 1 + 1 = 2a 1 + 7 a 2 = 2(5) + 7 a 2 = 17 a 2 + 1 = 2(17) + 7 a 3 = 41 a 3 + 1 = 2(41) + 7 a 4 = 89 a 4 + 1 = 2(89) + 7 a 5 = 185 5, 17, 41, 89, 185

5. 5 of 17 Algebra 2 Chapter 11 Sections 6 & 7Iteration Iteration is the process of composing a function with itself. For example start with x 0 = 3 and f(x) = 2x + 2 For example start with x 0 = 3 and f(x) = 2x + 2 Put x 0 in f(x 0 ) to get x 1, so x 1 = f(x 0 ) =2(3) + 2 = 8 x 2 = f(x 1 ) = f(8) = 18; x 3 = f(x 2 ) = f(18) = 38; and …

6. 6 of 17 Algebra 2 Chapter 11 Sections 6 & 7 Example 2 Find the first three iterates x 1, x 2, x 3 of the function f(x) = 3x – 1 for an initial value of x 0 = 5. x 1 = f(x 0 ) = 3(5) – 1 = 14 x 2 = f(x 1 ) = 3(14) – 1 = 41 x 3 = f(x 2 ) = 3(41) – 1 = 122

7. Chapter 11 Sec 7 The Binomial Theorem

8. 8 of 17 Algebra 2 Chapter 11 Sections 6 & 7 Pascal’s Triangle Use this to find the coefficients of powers on binomials. Before we start here are the ground rules for any expansion of the form (a + b) n. 1. There are n + 1 terms. 2. The exponent n of (a + b) n is the exponent of a in the first term and of b in the last term. 3. In successive terms, the exponent of a decreases by one and b increases by one. 4. The sum of the exponents in each term is n. 5. The coefficients are symmetric.

9. 9 of 17 Algebra 2 Chapter 11 Sections 6 & 7 Pascal’s Triangle 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 (a + b) 0 (a + b) 1 (a + b) 2 (a + b) 3 (a + b) 4 (a + b) 5 (a + b) 6 (a + b) 7

10. 10 of 17 Algebra 2 Chapter 11 Sections 6 & 7 Example 1 Expand (p + q) 5. (p + q) 5 = p 5 + 5p 4 q + 10p 3 q 2 + 10p 2 q 3 + 5pq 4 + q 5 1 5 10 10 5 1 (a + b) 5 1 4 6 4 1 (a + b) 4 Expand (x – 3) 4. (x – 3) 4 = x 4 + 4x 3 (–3) + 6x 2 (–3) 2 + 4x (–3) 3 + (–3) 4 = x 4 – 12x 3 + 54x 2 – 108x + 81 = x 4 – 12x 3 + 54x 2 – 108x + 81 1 4 6 4 1 (a + b) 4 Expand (3x + 2) 4. (3x + 2) 4 = (3x) 4 + 4(3x) 3 (2) + 6(3x) 2 (2) 2 + 4(3x) (2) 3 + (2) 4 = 81x 4 + 216x 3 + 216x 2 + 96x + 16 = 81x 4 + 216x 3 + 216x 2 + 96x + 16

11. 11 of 17 Algebra 2 Chapter 11 Sections 6 & 7 The Binomial Theorem Notice the rules still apply. 1. There are n + 1 terms. 2. The exponent n of (a + b) n is the exponent of a in the first term and of b in the last term. 3. In successive terms, the exponent of a decreases by one and b increases by one. 4.The sum of the exponents in each term is n. 5.The coefficients are symmetric.

12. 12 of 17 Algebra 2 Chapter 11 Sections 6 & 7 Using the B-Theorem Expand (t – s) 5. The expansion will have six terms. Use to find the first 3 terms, then use symmetry for the rest. (t – s) 5 = 1t 5 (–s) 0 +5t 4 (–s) 1 +10t 3 (–s) 2 +10t 2 (–s) 3 +5t(–s) 4 + (–s) 5 (t – s) 5 = 1t 5 (–s) 0 +5t 4 (–s) 1 +10t 3 (–s) 2 +10t 2 (–s) 3 +5t(–s) 4 + (–s) 5 = t 5 – 5t 4 s 1 + 10t 3 s 2 – 10t 2 s 3 + 5ts 4 – s 5 = t 5 – 5t 4 s 1 + 10t 3 s 2 – 10t 2 s 3 + 5ts 4 – s 5

13. 13 of 17 Algebra 2 Chapter 11 Sections 6 & 7Factorials The factors in the coefficients of the binomial theorem involve special products called factorials. For example the product is written 4! And is read 4 factorial. In general, if n is a positive integer, then By definition 0! = 1. Ex. Evaluate

14. 14 of 17 Algebra 2 Chapter 11 Sections 6 & 7 Binomial Theorem in Factorial Form Sometimes you need to know a particular term of a binomial expansion. Note that when the Binomial Theorem is written in Sigma notation, k = 0 for the first term, k = 1 for the second term, and so on. In general, the value of k is always one less than the term you are finding.

15. 15 of 17 Algebra 2 Chapter 11 Sections 6 & 7 Example 4 Expand (3x – y) 4. Now replace k with 0, 1, 2, 3, & 4 to get the five terms.

16. 16 of 17 Algebra 2 Chapter 11 Sections 6 & 7 Example 5 Find the fourth term in the expansion of (a + 3b) 4. Use the Binomial Theorem to write the expansion in sigma notation. n = 4, k = 3 (a + 3b) 4 =

17. 17 of 17 Algebra 2 Chapter 11 Sections 6 & 7 Daily Assignment Chapter 11 Sections 6 & 7 Study Guide (SG) Pg 151 – 152 Odd Pg 153 – 154 #1 – 5 on each page