Here's a simple handy little general proof of the result $\sum (x_i - \overline{x}) = 0$

Let's take the series of numbers: $$x_1,x_2,x_3,...,x_n$$ we acknowledge that the mean of this number set can be denoted by, $$\overline{x}=\frac{\sum x_i}{n}$$ Going back to the LHS of the original statement $\sum (x_i - \overline{x})$ we can write this out in full as follows: $$\sum (x_i - \overline{x}) = \Bigl(x_1-\frac{\sum x_i}{n}\Bigl) + \Bigl(x_2-\frac{\sum x_i}{n}\Bigl) + \Bigl(x_3-\frac{\sum x_i}{n}\Bigl) +...+\Bigl(x_n-\frac{\sum x_i}{n}\Bigl)$$ This can be simplified down to 0 in the following steps: $$x_1+x_2+x_3+...+x_n-\Bigl(\frac{n\sum x_i}{n}\Bigl)$$ $$\sum x_i-\sum x_i$$ $$=0$$

Here's a simple handy little general proof of the result $\sum (x_i - \overline{x}) = 0$

Let's take the sequence of numbers: $$x_1,x_2,x_3,...,x_n$$ we acknowledge that the mean of this number set can be denoted by, $$\overline{x}=\frac{\sum x_i}{n}$$ Going back to the LHS of the original statement $\sum (x_i - \overline{x})$ we can write this out in full as follows: $$\sum (x_i - \overline{x}) = \Bigl(x_1-\frac{\sum x_i}{n}\Bigl) + \Bigl(x_2-\frac{\sum x_i}{n}\Bigl) + \Bigl(x_3-\frac{\sum x_i}{n}\Bigl) +...+\Bigl(x_n-\frac{\sum x_i}{n}\Bigl)$$ This can be simplified down to 0 in the following steps: $$x_1+x_2+x_3+...+x_n-\Bigl(\frac{n\sum x_i}{n}\Bigl)$$ $$\sum x_i-\sum x_i$$ $$=0$$

Here's a simple handy little general proof of the result $\sum (x_i - \overline{x}) = 0$

Let's take the series of numbers: $$x_1,x_2,x_3,...,x_n$$ we acknowledge that the mean of this number set can be denoted by, $$\overline{x}=\frac{\sum x_i}{n}$$ Going back to the LHS of the original statement $\sum (x_i - \overline{x})$ we can write this out in full as follows: $$\sum (x_i - \overline{x}) = \Bigl(x_1-\frac{\sum x_i}{n}\Bigl) + \Bigl(x_2-\frac{\sum x_i}{n}\Bigl) + \Bigl(x_3-\frac{\sum x_i}{n}\Bigl) +...+\Bigl(x_n-\frac{\sum x_i}{n}\Bigl)$$ This can be simplified down to 0 in the following steps: $$x_1+x_2+x_3+...+x_4-\Bigl(\frac{n\sum x_i}{n}\Bigl)$$ $$\sum x_i-\sum x_i$$ $$=0$$

Here's a simple handy little general proof of the result $\sum (x_i - \overline{x}) = 0$

Let's take the series of numbers: $$x_1,x_2,x_3,...,x_n$$ we acknowledge that the mean of this number set can be denoted by, $$\overline{x}=\frac{\sum x_i}{n}$$ Going back to the LHS of the original statement $\sum (x_i - \overline{x})$ we can write this out in full as follows: $$\sum (x_i - \overline{x}) = \Bigl(x_1-\frac{\sum x_i}{n}\Bigl) + \Bigl(x_2-\frac{\sum x_i}{n}\Bigl) + \Bigl(x_3-\frac{\sum x_i}{n}\Bigl) +...+\Bigl(x_n-\frac{\sum x_i}{n}\Bigl)$$ This can be simplified down to 0 in the following steps: $$x_1+x_2+x_3+...+x_n-\Bigl(\frac{n\sum x_i}{n}\Bigl)$$ $$\sum x_i-\sum x_i$$ $$=0$$