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Unformatted text preview: EECS 203: Homework 4 Solutions Section 2.4 1. (E) 4cd c) a = 8 ,a 1 = 11 ,a 2 = 23 ,a 3 = 71 d) a = 2 ,a 1 = 0 ,a 2 = 8 ,a 3 = 0 2. (E) 10ef e) The general formula for this sequence is a n = (3 a n 1 ) + 2 , given that a = 0 Therefore, this sequence from n=0 to n=12 is: 0, 2, 8, 26, 80, 242, 728, 2186,6560, 19682, 59048 , 177146 , 531440 , ... f) The general formula for this sequence is a n = a n 1 (2 n + 1) , given that a = 1 Therefore, this sequence from n=0 to n=12 is: 1, 3, 15, 105, 945, 10345, 135135, 2027025, 34459425, 654729075 , 13749310575 , 316234143225 , ... 3. (E) 16cd c) There are two ways to answer the problem: using brute force, or using geometric series formula. i. Bruteforce method 8 X j =0 (2 3 j + 3 2 j ) = (2 3 + 3 2 ) + (2 3 1 + 3 2 1 ) + ... + (2 3 8 + 3 2 8 ) = 5 + 11 + ... + 13890 = 21215 ii. Geometricsum method 8 X j =0 (2 3 j + 3 2 j ) = 2 8 X j =0 3 j + 3 8 X j =0 2 j = 2 3 8+1 1 3 1 + 3 2 8+1 1 2 1 = 2 19682 2 + 3 511 1 = 21215 d) Like part (3c), you can solve the problem with either brute force, or geometric series formula. In addition, it is also possible to use telescoping to simplify the problem. i. Bruteforce method 8 X j =0 (2 j +1 2 j ) = (2 1 2 ) + (2 2 2 1 ) + ... + (2 9 2 8 ) = 1 + 2 + ... + 256 = 511 Note that the addition part from term a i can be canceled by the subtraction part from term a i +1 . This form of sum is called telescoping. ii. Geometricsum method 8 X j =0 (2 j +1 2 j ) = 2 8 X j =0 (2 j ) 8 X j =0 (2 j ) = 8 X j =0 (2 j ) = 2 8+1 1 2 1 = 511 1 = 511 iii. Telescoping method 8 X j =0 (2 j +1 2 j ) = (2 1 2 ) + (2 2 2 1 ) + ... + (2 9 2 8 ) = 2 9 2 = 511 The addition part from term a i can be canceled by the subtraction part from term a i +1 . This form of sum is called telescoping. Note that this technique only works on particular forms of summations. 4. (E) 18bd b) The double summation can be solved by either using brute force summation or by simplifying summa tions and then applying formulas. i. Bruteforce summation 3 X i =0 2 X j =0 (3 i + 2 j ) = (3(0) + 2(0)) + (3(1) + 2(0)) + (3(2) + 2(0)) + (3(3) + 2(0)) + ......
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This note was uploaded on 02/12/2012 for the course EECS 203 taught by Professor Yaoyunshi during the Spring '07 term at University of Michigan.
 Spring '07
 YaoyunShi

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