hw7-Sols - Homework 7 Solutions University of Pittsburgh...

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1 Homework 7 Solutions University of Pittsburgh Computer Science Department CS441 Discrete Structures for Computer Science Instructor: Milos Hauskrecht Chapter 5.1 4. Let P(n) be the proposition 1 3 + 2 3 + · · · + n 3 = (n(n + 1)/2) 2 whenever n is a positive integer. Basis step: We verify P(1), namely that 13 = [1 · (1 + 1)/2]2 which is indeed true. Next, Inductive step: We assume that P(k) is true (the inductive hypothesis), and try to derive P(k + 1). Now, P(k + 1) is the formula 1 3 + 2 3 + · · · + k 3 + (k+1) 3 = ((k + 1) (k + 2)/2) 2 Observe that all but the last term of the left-hand side (lhs) is exactly the lhs of P(k), so by the inductive hypothesis, it equals (k(k + 1)/2) 2 . Thus we have, 1 3 + 2 3 + · · · + k 3 + (k+1) 3 = ((k + 1) k/2) 2 +(k+1) 3 =(k+1) 2 ((k 2 /4)+k+1)= ((k + 1) (k + 2)/2) 2 7. Let P(n) be the proposition 3+3 · 5+3 · 5 2 · · ·+3 · 5 n = 3(5 n+1 −1)/4. To prove that P(n) is true for all nonnegative integers n, we proceed by mathematical induction. Basis step: We verify P(0), namely that 3 = 3(5−1)/4 which is indeed true. Next, Inductive step: We assume that P(k) is true (the inductive hypothesis), and try to derive P(k + 1). Now, P(k + 1) is the formula 3 + 3 · 5 + 3 · 5 2 · · · + 3 · 5 k + 3 · 5 K+1 = 3(5 K+2 − 1)/4 Observe that all but the last term of the left-hand side (lhs) is exactly the lhs of P(k), so by the inductive hypothesis, it equals 3(5 k+1 -1)/4. Thus we have, 3 + 3 · 5 + 3 · 5 2 · · · + 3 · 5 k + 3 · 5 k+1 =3(5 k+1 − 1)/4+ 3 · 5k+1 = 5 k+1 (3/4+ 3) −3/4=3(5 k+2 − 1)/4 10.
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