Answer
$-14$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To evaluate the given summation expression, $
\displaystyle\sum_{i=1}^3 (3x_i-x_i^2)
,$ substitute $
i
$ with the values from $
1
$ to $
3
.$ Then substitute the given values for each $x_i$'s
$\bf{\text{Solution Details:}}$
Substituting $
i
$ with the numbers from $
1
$ to $
3
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
(3x_1-x_1^2)+(3x_2-x_2^2)+(3x_3-x_3^2)
\\\\=
3x_1-x_1^2+3x_2-x_2^2+3x_3-x_3^2
\\\\=
(3x_1+3x_2+3x_3)+(-x_1^2-x_2^2-x_3^2)
\\\\=
3(x_1+x_2+x_3)-(x_1^2+x_2^2+x_3^2)
.\end{array}
Using the given values $x_1=-2,x_2=-1,x_3=0,x_4=1,$ and $x_5=2,$ the expression above evaluates to
\begin{array}{l}\require{cancel}
3((-2)+(-1)+0)-((-2)^2+(-1)^2+0^2)
\\\\=
3(-2-1+0)-(4+1+0)
\\\\=
3(-3)-5
\\\\=
-9-5
\\\\=
-14
.\end{array}
