Answer
$\sum_{i=1}^{n}\left(-\bar{x}+x_{i}\right)=-\sum_{i=1}^{n} \bar{x}+\sum_{i=1}^{n} x_{i}=\sum_{i=1}^{n}-\frac{\sum_{i=1}^{n} x_{i}}{n} \cdot n +x_{i}=0$
Work Step by Step
Let $ \bar {x} $ be the average of the numbers $x_{1}, x_{2}, \ldots, x_{n}$
\[
\therefore \bar{x}=\frac{\sum_{i=1}^{n} x_{i}}{n}
\]
Theorem ( $5.4 .1)$
\[
\begin{aligned}
\sum_{i=1}^{n}\left(-\bar{x}+x_{i}\right) &=\sum_{i=1}^{n} x_{i}-\sum_{i=1}^{n} \bar{x} \\
&=-\bar{x} \sum_{i=1}^{n} 1+ \sum_{i=1}^{n} x_{i}\\
&=\sum_{i=1}^{n} x_{i}-\frac{\sum_{i=1}^{n} x_{i}}{n} \cdot n \\
&=-\sum_{i=1}^{n} x_{i}+\sum_{i=1}^{n} x_{i} \\
&=0
\end{aligned}
\]
