Sequences and Series, Induction. Review

Transcription

1 Sequences and Series, Induction Review 1

2 Topics Arithmetic Sequences Arithmetic Series Geometric Sequences Geometric Series Factorial Notation Sigma Notation Binomial Theorem Mathematical Induction 2

3 Arithmetic Sequence Sequence of numbers The difference between any two sequential terms is the same Examples: 1, 3, 5, 7, 9 5, 10, 15, 20, 25 Can write as a, a+d, a+2d, a+nd where a is the first term and d is the common difference The nth term is a+ (n-1)d 3

4 Example The fifth term in an arithmetic sequence is ½ and the 20 th is 7/8. Find the first three terms of the sequence. 4

5 Solution The fifth term in an arithmetic sequence is ½ and the 20 th is 7/8. Find the first three terms of the sequence. a + (5-1)d = ½, where a is the 1 st term, and d is the common difference a + (20-1)d = 7/8 Taking the 2 nd equation less the 1 st, we have 15d = 3/8, or d = 1/40 a+4/40 = ½, from the equation for the fifth term, so a = 2/5. Check: 2/5 + 1/10 = 5/10 = ½ 2/5 + 19/40 = 16/ /40 = 35/40 = 7/8 The first three terms are 2/5, 17/40, 9/20 5

6 Example Find the common difference in an arithmetic sequence in which the 10 th term minus the 20 th is 70 6

7 Solution Find the common difference in an arithmetic sequence in which the 10 th term minus the 20 th is 70 Note, that since the 20 th is 70 less then 10 th, the difference must be negative (a + 9d) (a + 19d) = 70-10d = 70, d = -7 7

8 Arithmetic Series The sum of an arithmetic sequence Assume a n = a + (n-1)d, where d is the common difference and a is the first term The sum of the first n terms is n(a + a n )/2 8

9 Examples Find the sum of the first 16 terms in the sequence 2, 11, 20, 29 Find the sum of the first 50 terms in a series whose first term is -8 and 50 th term is 139 9

10 Solution Find the sum of the first 16 terms in the sequence 2, 11, 20, 29 difference is 9, a n = 2+(n-1)9; n=16 so a n = 2+(15)9=137 so sum is 16[1/2(2+137)] = 1112 Find the sum of the first 50 terms in a series whose first term is -8 and 50 th term is ( )/2 = 25(131) =

11 Example The sum of the first 12 terms of an arithmetic sequence is 156. What is the sum of the 1 st and 12 th term? 11

12 Solution The sum of an arithmetic sequence of length n is n times the mean of the 1 st and nth term The sum of the first 12 terms of an arithmetic sequence is 156. What is the sum of the 1 st and 12 th term? 156 = 12(a1 + a12)/2 a1 + a12 = 312/12 = 26 12

13 Geometric Sequences and Series A geometric sequence (or progression) is a sequence of the form a, ar, ar 2, ar 3,, where a and r are non-zero constants (a and r can be < 0) r is called the common ratio For example, if we have a geometric series 3, 6, 12, then a = 3 and r = 2 13

14 Finding the nth term a n = ar n-1 Find the 7th term in the sequence 2, 6, 18,... a = 2, and r = 3, the 7 th term is 2(3 6 ) = 2(729) = 1458 As with arithmetic sequences, we can have a finite geometric sequences 14

15 Examples Find the 100 th term of the sequence -1, 1, -1, 1 Find the 6 th term of the series 1, -2 2, 8 15

16 Solution Find the 100 th term of the series -1, 1, -1, 1 The ratio is -1, so we have -1(-1) 99 = 1 Find the 6 th term of the series 1, - 2, 2 Ratio is - 2, so we have 1(- 2) n-1 = 1(- 2) 5 =

17 A Finite Geometric Series a + ar + ar 2 + ar 3 +ar n-1 = a(1 rn ) 1 r 17

18 Example Find the sum of 1/2 + (1/2) 2 + (1/2) 9 18

19 Solution Find the sum of 1/2 + (1/2) 2 + (1/2) 9 Sum = (1 1 ) = 1 2 ( ) = 1-1/1024 = 1023/

20 An Infinite Geometric Series Consider ½ + 1/4 + 1/8 + For n terms, the sum is n = (1- ( ½ ) n ) As n gets large, this is just 1 The sum of an infinite series, with r <1 is a 1 r 20

21 Example 2 Find k=1 3 k 1 21

22 Solution 2 Find k=1 3 k 1 s = a = 1 = 3 1 r 1/3 22

23 Example Find the sum of the infinite series 9/10 + 9/ /

24 Solution Find the sum of the infinite series 9/10 + 9/ /1000 Sum = a/(1-r) = (9/10) / ((1-1/10) = 1 24

25 Repeating Decimals Find a fractional equivalent of = 2/ / /100,000 Treat as an infinite geometric series Sum = 2/10 + a(1-r) = 2/10 + (35/1000)/(1-1/100) =2/ /( 1000(99/100)) = 2/ /(990)= 233/990 25

26 Evaluate Example 26

27 Solution Evaluate a=47/100 r = 1/100 S = a/(1-r) = (47/100)/(99/100) = 47/99 27

28 Sigma Notation 28

29 Sigma notation: A Shortcut Notation m n=k a n Means to sum the values a, starting with n = k and ending with n = m, n=k m a n = a k +a k+1 + a m 29

30 Example 6 j=4 2 j =

31 Example x + x2 + x3 + x4 1! 2! 3! 4! x12 = 12 x j 12! j=1 j! 31

32 Examples 5 Find k=1 k 32

33 Solution 5 Find k=1 k Sum is n( a1 + an)/2 = 5(1+5)/2 = 15 33

34 Factorial n! = 1 x 2 x 3 x 4 x n 1! = 1 0! = 1 (by definition) 34

35 Example Find (n+1)!/(n-1)! = n(n+1)= n 2 + n 35

36 Binomial Theorem The value of (a+b) n can (a+b) = a + b (a+b) 2 = a 2 + 2b + b 2 (a+b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 (a+b) 4 = a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4 Note the pattern of the coefficients 36

37 n k = n! for n k k! n k! The Binomial Coefficient 5 3 = 5! 3!2! = 5 4 = 10 (3 2 1)(2 1)

38 The Binomial Theorem (a+b) n = n 0 an + n 1 an-1 b + n 2 an-2 b 2 + n n 1 abn-1 + n n bn n k = n! k! n k! for n k So, (a+b) 3 = 3! 0!3! a3 + 3! 1!2! a2 b+ 3! 2!1! ab2 + 3! 3!0! b3 = a 3 + 3a 2 b+ 3ab 2 + b 3 38

39 General Expansion The rth term in the expansion of (a+b) n is: n r 1 an-r+1 b r-1 39

40 Practice Find (a+b) 6 40

41 Solution Find (a+b) 6 a 6 + 6a 5 b + 15a 4 b a 3 b a 2 b 4 + 6ab 5 + b 6 41

42 Find the 5 th term of (a-2b) 3 Example 42

43 Find the 3rd term of (a-2b) 5 Solution 5! 2! 5 2! a3 (2b) 2 = a 3 (b) 2 = 40a 3 b

44 Mathematical Induction Prove the statement is true for k = 1 Prove that if the statement is true for k, then it is true for k+1 44

45 Example (2n-1) = n 2 Let n = 1: (2n-1) = 1= n 2 Assume (2n-1) = n 2 Add the n+1 term to both sides Show n 2 + 2(n + 1) = (n+1) 2 =n 2 + 2n + 2 = n 2 + 2(n +1) Expression remains true 45

46 Example Show (2n) 3 = 2n 2 (n+1) 2 46

47 Solution Show (2n) 3 = 2n 2 (n+1) 2 n=1: (2) 3 = 2(1) 3 (1+1) 2 or 8 = 2(4) = 8 n+1: Add term to left side (2(n+1)) 3 gives (2n) 3 +(2(n+1)) 3 Add to right side; 2n 2 (n+1) 2 + [2(n + 1)] 3 = 2n 2 (n+1) 2 + 8(n+ 1) 3 = 2(n+1) 2 [ n 2 + 4(n+1)]= 2(n+1) 2 (n+2) 2 47