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Unformatted text preview: Chapter 5: Sequences and Mathematical Induction Section 5.1 6. f 1 = „ 1 4 ” · 4 = 0 · 4 = 0, f 2 = „ 2 4 ” · 4 = 0 · 4 = 0, f 3 = „ 3 4 ” · 4 = 0 · 4 = 4 , f 4 = „ 4 4 ” · 4 = 1 · 4 = 4 9. h 1 = 1 · b log 2 1 c = 1 · h 2 = 2 · b log 2 2 c = 2 · 1 h 3 = 3 · b log 2 3 c = 3 · 1 h 4 = 4 · b log 2 4 c = 4 · 2 h 5 = 5 · b log 2 5 c = 5 · 2 h 6 = 6 · b log 2 6 c = 6 · 2 h 7 = 7 · b log 2 7 c = 7 · 2 h 8 = 8 · b log 2 8 c = 8 · 3 h 9 = 9 · b log 2 9 c = 9 · 3 h 10 = 10 · b log 2 10 c = 10 · 3 h 11 = 11 · b log 2 11 c = 11 · 3 h 12 = 12 · b log 2 12 c = 12 · 3 h 13 = 13 · b log 2 13 c = 13 · 3 h 14 = 14 · b log 2 14 c = 14 · 3 h 15 = 15 · b log 2 15 c = 15 · 3 When n is an integral power of 2, h n is n times the exponent of that power. For instance, 8 = 2 3 and h 8 = 8 · 3. If m and n are integers and 2 m ≤ n < 2 m +1 , then h n = n · m . 15. a n = ( 1) n 1 n 1 n ¶ for all integers n ≥ 1 (There are other correct answers for this exercise.) 18. e . Q 2 k =2 a k = a 2 = 2 21. 3 X m =0 1 2 m = 1 2 + 1 2 1 + 1 2 2 + 1 2 3 = 1 + 1 2 + 1 4 + 1 8 = 15 8 24. X j =0 ( j + 1) · 2 j = (0 + 1) · 2 = 1 · 1 = 1 30. n X j =1 j ( j + 1) =1 · 2 + 2 · 3 + 3 · 4 + ··· + n · ( n + 1) (There are other correct answers for this exercise.) 36. 1 · 2 3 · 4 ¶ = 1 6 39. n +1 X m =1 m ( m + 1) = n X m =1 m ( m + 1) + ( n + 1)(( n + 1) + 1) 42. n X m =0 ( m + 1)2 m + ( n + 2)2 n +1 = n +1 X m =0 ( m + 1)2 m Exercises 45 and 48 have more than one correct answer. 38 Chapter 5: Sequences and Mathematical Induction 45. 4 Y i =2 ( i 2 1) 48. 4 Y j =1 (1 t j ) 54. When k = 1, i = 1 + 1 = 2. When k = n , i = n + 1. Since i = k + 1, then k = i 1. So k k 2 +4 = ( i 1) ( i 1) 2 +4 = i 1 i 2 2 i + 1+4 = i 1 i 2 2 i + 5 . Therefore, n Y k =1 ( k k 2 +4 ) = n +1 Y i =2 ( i 1 i 2 2 i + 5 ) . 57. When i = 1, j = 1 1 = 0. When i = n 1, j = n 2. Since j = i 1, then i = j + 1. So i ( n i ) 2 = j + 1 ( n ( j + 1)) 2 = j + 1 ( n j 1) 2 . Therefore, n 1 X i =1 ( i ( n i ) 2 ) = n 2 X j =0 ( j + 1 ( n j 1) 2 ) . 60. By Theorem 5.1.1, 2 · n X k =1 (3 k 2 + 4) + 5 · n X k =1 (2 k 2 1) = n X k =1 2(3 k 2 + 4) + n X k =1 5(2 k 2 1) = n X k =1 (6 k 2 + 8) + n X k =1 (10 k 2 5) = n X k =1 (6 k 2 + 8 + 10 k 2 5) = n X k =1 (16 k 2 + 3) . 63. 6! 8! = 6! 8 · 7 · 6! = 1 56 72. 7 4 ¶ = 7! 4!(7 4)! = 7 · 6 · 5 · 4 · 3 · 2 · 1 (4 · 3 · 2 · 1)(3 · 2 · 1) = 7 · 6 · 5 (3 · 2 · 1) = 35 87. Let a nonnegative integer a be given. Divide a by 16 using the quotientremainder theorem to obtain a quotient q [0] and a remainder r [0]. If the quotient is nonzero, divide by 16 again to obtain a quotient q [1] and a remainder r [1]. Continue this process until a quotient of 0 is obtained. The remainders calculated in this way are the hexadecimal digits of a : a 10 = ( r [ k ] r [ k 1] ...r [2] r [1] r [0]) 16 . Section 5.2 39 90....
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This note was uploaded on 05/10/2011 for the course MATH 325K taught by Professor Shirley during the Spring '11 term at University of Texas.
 Spring '11
 Shirley
 Mathematical Induction

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