Homework 5 Solutions

Homework 5 Solutions - EECS 203 Homework 5 Solutions...

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EECS 203: Homework 5 Solutions Section 2.4 1. (E) 16bc b) j = 0 8 3 j 2 j = j = 0 8 3 j j = 0 8 2 j = 3 9 1 3 1 2 9 1 2 1 = 9841 511 = 9330 c) j = 0 8 2 3 j 3 2 j = 2 j = 0 8 3 j 3 j = 0 8 2 j = 2 3 9 1 3 1 3 2 9 1 2 1 = 2 9841 3 511 = 21215 2. (E) i = 1 n j = 1 m 2 i 3 j = i = 1 n j = 1 m 2 i i = 1 n j = 1 m 3 j = i = 1 n 2 mi i = 1 n 3 m ⋅ m 1 2 = 2 m n ⋅ n 1 2 3 n m ⋅ m 1 2 = n m ⋅ 2 n 3 m 5 2 3. (E) j = 1 m i = 1 n 2 i 3 j = j = 1 m i = 1 n 2 i j = 1 m i = 1 n 3 j = j = 1 m 2 n ⋅ n 1 2 j = 1 m 3 n j = 2 m n ⋅ n 1 2 3 n m ⋅ m 1 2 = n m ⋅ 2 n 3 m 5 2
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4. (M) 26 Solution 1: The problem is to find a formula for k = 0 m 3 k . The sum begins with (1+1+1+1+1+1+1)+(2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2)+(3+3+. ..+3)+. ... The twos start with term 8. The threes start with term 27. In general, the number i starts with term i 3 . Consider subtracting one out of each element of the term. Then a total of m is subtracted, and the remaining terms are:
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Homework 5 Solutions - EECS 203 Homework 5 Solutions...

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