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Unformatted text preview: Math 326 Fall 2010 Homework Set Chapter 4 Solutions 1 4.17 Let a k = 2 k + 1 adn b k = ( k 1) 3 + k + 2 for all integers k ≥ 0. Show that the first three terms of these sequences are identical but that their 4th terms differ. a = 2 · 0 + 1 = 1 a 1 = 2 · 1 + 1 = 3 a 2 = 2 · 2 + 1 = 5 a 3 = 2 · 3 + 1 = 7 . . . b = (0 1) 3 + 0 + 2 = ( 1) 3 + 2 = 1 b 1 = (1 1) 3 + 1 + 2 = (0) 3 + 3 = 3 b 2 = (2 1) 3 + 2 + 2 = (1) 3 + 4 = 5 b 3 = (3 1) 3 + 3 + 2 = (2) 3 + 5 = 13 . . . 4.113 Find an explicit formula for the sequence: 1 1 2 , 1 2 1 3 , 1 3 1 4 , 1 4 1 5 , 1 5 1 6 , 1 6 1 7 ,... a k = 1 k 1 k + 1 ,k = 1 , 2 ,... 4.116 Find an explicit formula for the sequence: 3 , 6 , 12 , 24 , 48 , 96 ,... a k = 3 · 2 k , k = 0 , 1 , 2 ,... 4.122 Compute the summation 4 Y j =0 ( 1) j 4 Y j =0 ( 1) j = ( 1) ( 1) 1 ( 1) 2 ( 1) 3 ( 1) 4 = 1 4.133 (1 3 1) (2 3 1) + (3 3 1) (4 3 1) + (5 3 1) = 5 X k =1 ( 1) k +1 ( k 3 1) 4.134 (2 2 1)(3 2 1)(4 2 1) = 4 Y k =2 ( k 2 1) 4.137 (1 t )(1 t 2 )(1 t 3 )(1 t 4 ) = 4 Y k =1 (1 t k ) 4.139 1 2! + 2 3! + ··· + n ( n + 1)! = n X i =2 k ( k + 1)! 4.153 Change variables i = k + 1 in n Y k =1 k k 2 + 4 n Y k =1 k k 2 + 4 = n +1 Y i =1+1 i 1 ( i 1) 2 + 4 2 4.157 Change the variables using j = i 1 in 2 n Y i = n n i + 1 n + i 2 n Y i = n n i + 1 n + i = 2 n 1 Y j = n 1 n ( j + 1) + 1 n + j + 1 = 2 n 1 Y j = n 1 n j n + j + 1 4.159 Write as a single sum: 2 n X k =1 (3 k 2 + 4) + 5 n X k =1 (2 k 2 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) + 5(2 k 2 1) = n X k =1 6 k 2 + 8 + 10 k 2 5 = n X k =1 16 k 2 + 3 4.162 Suppose a[1],a[2],...,a[m] is a one dimensional array and consider the following algorithm segment: sum = 0; for k=1 to m sum = sum + a[k] next k Fill in the blanks below so that each algorithm segment performs the same job as the one given above. (a) sum = 0; for i=0 to ____ sum = _____ next i answer: m 1; sum + a[i+1] (b) sum = 0; for j=2 to ____ sum = _____ next j answer: m + 1; sum + a[j1] 4.169 Write an informal description of an algorithm using repeated division by 16 to convert a nonnegative integer from decimal notation to hexadecimal. Let a nonnegative integer a be given. Divide a by 16 using the quotient remainder theorem to obtain a quotient q[0] and a re mainder r[0] . If the quotient is nonzero, divide by 16 to obtain a quotient q[1] and a remainder r[1] . 3 Continue this process until a quotient of 0 is obtained. The sequence of remainders is the list of hexadecimal digits of a . 4.173 Write a formal version of the algorithm you developed in exercise 69. Input: a Algorithm Body: q=a; i=0; while i=0 or q != 0, r[i]= q mod 16 q = q div 16 i = i + 1 end while Output: r[0], r[1], r[2], ..., r[i1] End Algorithm 4 4.22 Use mathematical induction to show that any postage of at least 12 cents can be obtained using the 3 cent and 7 cent stamps....
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This note was uploaded on 06/12/2011 for the course MATH 103 taught by Professor Wouters during the Spring '08 term at Wisc Oshkosh.
 Spring '08
 WOUTERS
 Algebra, Integers

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