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Homework 4 Solutions W11

# Homework 4 Solutions W11 - EECS 203 Homework 4 Solutions...

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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. Brute-force 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. Geometric-sum 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. Brute-force 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. Geometric-sum 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. Brute-force 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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Homework 4 Solutions W11 - EECS 203 Homework 4 Solutions...

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