Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 5 - Sequences, Mathematical Induction, and Recursion - Exercise Set 5.2 - Page 256: 4

Answer

(a) $1(1+1)=\frac{2(2-1)(2+1)}{3}$, true. (b) $\sum_{i=1}^{k-1}i(i+1)=\frac{k(k-1)(k+1)}{3}$ (c) $\sum_{i=1}^{k}i(i+1)=\frac{(k+1)k(k+2)}{3}$ (d) if P(k) is true then P(k+1) will also be true

Work Step by Step

(a) For $n=2$, we have $LHS=1(1+1)=2$ and $RHS=\frac{2(2-1)(2+1)}{3}=2$, thus P(2) is true. (b) For $P(k)$, we have $\sum_{i=1}^{k-1}i(i+1)=\frac{k(k-1)(k+1)}{3}$ (c) For $P(k+1)$, we have $\sum_{i=1}^{k}i(i+1)=\frac{(k+1)k(k+2)}{3}$ (d) To prove by mathematical induction that the formula holds for all integers $n\ge2$, we must show that if P(k) is true then P(k+1) will also be true in the inductive step.
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