College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 8 - Sequences, Induction, and Probability - Exercise Set 8.4 - Page 751: 14


See below.

Work Step by Step

Proofs using mathematical induction consists of two steps: 1) The base case: here we prove that the statement holds for the first natural number. 2) The inductive step: assume the statement is true for some arbitrary natural number $n$ larger than the first natural number, then we prove that then the statement also holds for $n + 1$. Hence here: 1) For $n=1: 3=\frac{3(1)(1+1)}{2}$. 2) Assume for $n=k: 3+6+...+3k=\frac{3(k)(k+1)}{2}$. Then for $n=k+1$: $3+6+...+3k+3k+3=\frac{3(k)(k+1)}{2}+3k+3=\frac{3k^2+3k}{2}+\frac{6k+6}{2}=\frac{3k^2+9k+6}{2}=\frac{3(k+1)(k+2)}{2}=\frac{3(k+1)((k+1)+1)}{2}.$ Thus we proved what we wanted to.
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