Precalculus (6th Edition) Blitzer

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

Chapter 10 - Review Exercises - Page 1125: 57


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Work Step by Step

$\begin{align} & {{S}_{1}}:5=\frac{5\left( 1 \right)\left( 1+1 \right)}{2} \\ & 5=\frac{5\left( 2 \right)}{2} \\ & 5=5\ \text{is}\ \text{true} \end{align}$ $\begin{align} & {{S}_{k}}=5+10+15+....+5k=\frac{5k\left( k+1 \right)}{2} \\ & {{S}_{k+1}}=5+10+15+....+5k+5\left( k+1 \right) \\ & {{S}_{k+1}}=\frac{5\left( k+1 \right)\left( k+2 \right)}{2} \end{align}$ And add $5\left( k+1 \right)$ to both sides of ${{S}_{k}}$ $5+10+15+....+5k+5\left( k+1 \right)=\frac{5k\left( k+1 \right)}{2}+5k\left( k+1 \right)$ Then, simplify the right-hand side: $\begin{align} & \frac{5k\left( k+1 \right)}{2}+5\left( k+1 \right)=\frac{5k\left( k+1 \right)+10\left( k+1 \right)}{2} \\ & \frac{5k\left( k+1 \right)}{2}+5\left( k+1 \right)=\frac{\left( 5k+10 \right)\left( k+1 \right)}{2} \\ & \frac{5k\left( k+1 \right)}{2}+5\left( k+1 \right)=\frac{5\left( k+1 \right)\left( k+2 \right)}{2} \\ \end{align}$ When ${{S}_{k}}$ is true then ${{S}_{k+1}}$ is true.
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