Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 11 - Review Exercises - Page 841: 66


See below.

Work Step by Step

Proofs using mathematical induction consist 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 $n=1: 1=\frac{1}{4}(1+3)$ 2) Assume for $n=k: 1+1.5+...+0.5(k+1)=\frac{k}{4}(k+3)$. Then $n=k+1:1+1.5+...+0.5(k+1)+0.5(k+2)=\frac{k}{4}(k+3)+0.5(k+2)=0.25k^2+0.75k+0.5k+1=0.25(k+1)(k+4)=0.25(k+1)((k+1)+3)$ Thus we proved what we wanted to.
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