Homework_4.pdf

# Homework_4.pdf - Homework 4 ayush.parikh ID 112228873 March...

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Homework 4 ayush.parikh ID: 112228873 March 2019 Page 183 #11 a n = ( - 1) n n/(n+1) #21 4+3+4 = 11 #39 5 n =0 ( - 1) n +1 ( n 3 - 1) #53 Q 2 n - 1 j = n - 1 (n-j)/(n+1) #56 Q n k =1 ( k 2 + k ) / ( k 2 + 3 k + 2) Page 197 #4 a) 1(1+1) = 2 P(2) = 2(2-1)(2+1)/3 = 2 2 = 2, thus P(2) is true. b) P(k) = k(k-1)(k+1)/3 c) P(k+1) = (k+1)(k)(k+2)/3 d) In the inductive step, you have to assume that the summation is equal to P(k) for k 2 #9 1

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Base Case: P(3) = 4(4 3 - 16) / 3 = 4(48) / 3 = 192 / 3 = 64 = 4 3 Thus P(3) is true because 4 3 = 4 3 Inductive Step: Assume that P(k) is true for k geq 3 P(k) = 4 3 + 4 4 + 4 5 + ... + 4 k = 4(4 k - 16) / 3 (inductive hypothesis) We must show that P(k+1) is true for k 3 P(k+1) = 4 3 + 4 4 + 4 5 + ... + 4 k + 4 k +1 = 4(4 k +1 - 16) / 3 Taking the left hand side of P(k+1): 4 3 + 4 4 + 4 5 + ... + 4 k + 4 k +1 = (4(4 k - 16) / 3) + (4 k +1 ) by inductive hypothesis = (4 k +2 - 64) / 3 = 4(4 k +1 - 16) / 3 This is equal to the right side of the equation, so P(k+1) is true. #16 Base Case: (2+1)/(2*2) = 3/4 1 - 1/4 = 3/4 Thus P(2) is true since 3/4 = 3/4 Inductive Step: Assume that P(k) is true for k 2 P(k) = (1 - 1 / 2 2 )(1 - 1 / 3 2 ) ... (1 - 1 /k 2 ) = (k+1)/2k (inductive hypothesis) We must show that P(k+1) is true for k 2
• Fall '19
• Prime number, 3k, 2k, 4k, 5k, 1 J

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