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 257: 13

Answer

See below.

Work Step by Step

1. For $n=2$, we have $LHS=1(1+1)=2$ and $RHS=\frac{2(2-1)(2+1)}{3}=2=LHS$, thus P(2) is true. 2. Assume P(k) ($k\gt2$) is true, we have: $\sum_{i=1}^{k-1}=\frac{k(k-1)(k+1)}{3}$ 3. For $n=k+1$, we have: $LHS=\sum_{i=1}^{k-1}+k(k+1)=\frac{k(k-1)(k+1)}{3}+k(k+1)\\ =\frac{k(k-1)(k+1)+3k(k+1)}{3}=\frac{k(k+1)(k-1+3)}{3}=\frac{k(k+1)(k+2)}{3}=RHS$ 4. Thus $P(k+1)$ is also true and we have proved the statement by mathematical induction.
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