## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 11 - 11.4 - Mathematical Induction - 11.4 Exercises - Page 806: 14

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 $n=1: 1=\frac{1}{2}(3(1)-1)$ 2) Assume for $n=k: 1+4+...+(3k-2)=\frac{k}{2}(3(k)-1)$. Then for $n=k+1:1+4+...+(3k-2)+(3k+1)=\frac{k}{2}(3(k)-1)+(3k+1)=1.5k^2-0.5k+3k+1=1.5k^2+2.5k+1=\frac{k+1}{2}(3k+2)=\frac{k+1}{2}(3(k+1)-1)$ Thus we proved what we wanted to.

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