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: 12

Answer

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 $n=1: 3=\frac{1(1+5)}{2}$. 2) Assume for $n=k: 3+4+...+k+2=\frac{k(k+5)}{2}$. Then for $n=k+1$: $3+4+...+k+2+k+3=\frac{k(k+5)}{2}+k+3=\frac{k^2+5k}{2}+\frac{2k+6}{2}=\frac{k^2+7k+6}{2}=\frac{(k+1)(k+6)}{2}=\frac{(k+1)((k+5)+1)}{2}.$ Thus we proved what we wanted to.
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