Answer
$-10-4+0$
$=-14$
Work Step by Step
$\sum_{i=1}^{3}(3x_i-x_i^2)$
$=(3x_1-x_1^2)+(3x_2-x_2^2)+(3x_3-x_3^2)$
$=(3*(-2)-(-2)^2)+(3*(-1)-(-1)^2)+(3*0-0^2)$
$=-10-4+0$
$=\boxed{-14}$
Precalculus (6th Edition)
by Lial, Margaret L.; Hornsby, John; Schneider, David I.; Daniels, Callie
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
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