Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Test - Page 1127: 16


See explanations.

Work Step by Step

Step 1. Test for $n=1$, $LHS=1$, $RHS=\frac{1(3(1)-1)}{2}=1$. We have $LHS=RHS$, so it is true for $n=1$ Step 2. Assume it is true for $n=k$. We have $1+4+7++(3k-2)=\frac{k(3k-1)}{2}$ Step 3. For $n=k+1$, we have $LHS=1+4+7++(3k-2)+(3(k+1)-2)=\frac{k(3k-1)}{2}+(3k+1)=\frac{1}{2}(3k^2-k+6k+2)=\frac{1}{2}(3k^2+5k+2)=\frac{1}{2}(k+1)(3k+2)$ Step 4. For $n=k+1$, we have $RHS=\frac{1}{2}(k+1)(3(k+1)-1)=\frac{1}{2}(k+1)(3k+2)$ Step 5. We see that $LHS=RHS$ for $n=k+1$, assuming it is true for $n=k$. Thus, we proved the formula through mathematical induction.
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