Problem

Hello, tahsocks!

I'll re-write the problem in a more manageable form.

Quote:

$\text{Show that: }\:\sum^n_{i=1}\!x_i(x_i- \overline{x}) \;=\;\sum^n_{i=1}\!\left(x_i - \overline {x})^2$

Recall that $\overline{x}$ is a constant: . $\overline{x} \:=\:\frac{\sum x_i}{n} \quad\Rightarrow\quad \sum\! x_i \:=\:n\!\cdot\!\overline{x}$

The left side is: . $\sum\! x_i(x_i - \overline{x}) \;\;=\;\;\sum \!(x_1^2 - \overline{x}\,x_i)$

. . $=\;\;\sum\! x_i^2 - \overline{x}\sum x_i \;\;=\;\;\sum\! x_i^2 - \overline{x}\left(n\!\cdot\!\overline{x}\right) \;\;=\;\;\sum\!x_1^2 - n\!\cdot\!\overline{x}^2$

The right side is: . $\sum\!(x_1 - \overline{x})^2 \;\;=\;\;\sum\left(x_i^2 - 2\overline{x}x_i + \overline{x}^2\right)$

. . $=\;\;\sum\!x_i^2 - 2\overline{x}\sum\!x_i + \overline{x}^2\sum\!1 \;\;=\;\; \sum\!x_i^2 - 2\overline{x}(n\!\cdot\!\overline{x}) + \overline{x}^2\!\cdot\!n$

. . $=\;\; \sum\! x_i^2 - 2n\!\cdot\!\overline{x}^2 + n\!\cdot\!\overline{x}^2 \;\;=\;\; \sum\!x_i^2 - n\!\cdot\!\overline{x}^2$

The left side and the right side are equal . . . $\tex{ta-}D\!AA!$