Precalculus (6th Edition) Blitzer

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

Chapter 10 - Section 10.3 - Geoetric Sequences and Series - Exercise Set - Page 1077: 118


$$\frac{\left( k\text{ }+\text{ }1 \right)\left( k+2 \right)\left( 2k+3 \right)}{6}$$.

Work Step by Step

The term is $\frac{k\left( k\text{ }+\text{ }1 \right)\left( 2k\text{ }+\text{ }1 \right)}{6}+{{\left( k\text{ }+\text{ }1 \right)}^{2}}$. Taking out the $\left( k\text{ }+\text{ }1 \right)$ common term, we get, $\begin{align} & \frac{k\left( k\text{ }+\text{ }1 \right)\left( 2k\text{ }+\text{ }1 \right)}{6}+{{\left( k\text{ }+\text{ }1 \right)}^{2}}=\left[ \text{ }\frac{k\left( 2k\text{ }+\text{ }1 \right)}{6}+\left( k\text{ }+\text{ }1 \right) \right]\left( k\text{ }+\text{ }1 \right) \\ & =\frac{\left[ \text{ 2}{{k}^{2}}+k+6k+6 \right]}{6}\left( k\text{ }+\text{ }1 \right) \\ & =\frac{\left[ \text{ 2}{{k}^{2}}+7k+6 \right]}{6}\left( k\text{ }+\text{ }1 \right) \\ & =\frac{\left( k\text{ }+\text{ }1 \right)\left( k+2 \right)\left( 2k+3 \right)}{6} \end{align}$
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