College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Section 8.5 - Mathematical Induction - 8.5 Exercises - Page 628: 7

Answer

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