Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 11 - Further Topics in Algebra - 11.1 Sequences and Series - 11.1 Exercises - Page 1015: 63


$-10-4+0$ $=-14$

Work Step by Step

$\sum_{i=1}^{3}(3x_i-x_i^2)$ $=(3x_1-x_1^2)+(3x_2-x_2^2)+(3x_3-x_3^2)$ $=(3*(-2)-(-2)^2)+(3*(-1)-(-1)^2)+(3*0-0^2)$ $=-10-4+0$ $=\boxed{-14}$
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