Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 14 - Sequences, Series, and the Binomial Theorem - 14.1 Sequences and Series - 14.1 Exercise Set - Page 895: 63

Answer

$\sum\limits_{k=0}^{5}{\left( {{k}^{2}}-2k+3 \right)}$ is $\text{43}\text{.}$

Work Step by Step

$\sum\limits_{k=0}^{5}{\left( {{k}^{2}}-2k+3 \right)}$ For the sum of the notation, $\begin{align} & \sum\limits_{k=0}^{5}{\left( {{k}^{2}}-2k+3 \right)}=\left( {{0}^{2}}-2\cdot 0+3 \right)+\left( {{1}^{2}}-2\cdot 1+3 \right)+\left( {{2}^{2}}-2\cdot 2+3 \right)+\left( {{3}^{2}}-2\cdot 3+3 \right)+\left( {{4}^{2}}-2\cdot 4+3 \right)+\left( {{5}^{2}}-2\cdot 5+3 \right) \\ & =3+2+3+6+11+18 \\ & =43 \end{align}$ Thus, the sum of the sigma notation $\sum\limits_{k=0}^{5}{\left( {{k}^{2}}-2k+3 \right)}$ is $\text{43}\text{.}$
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