Derivation of $\sigma^2 = \frac1N\sum x^2 - \frac1{N^2}\sum\bar{x}^2$

Question

I saw the above equation in an introductory statistics textbook, as a shortcut for evaluating the variance of a population.

I tried to prove it myself:
 
 
$$\sigma^2 = \frac{\sum (x - \bar{x})^2}{N} \tag{1}$$ $$\sigma^2 = \frac{\sum x^2 - \frac{(\sum x)^2}{N}}{N} \tag{2}$$ We are given that $(1) = (2)$: $$\frac{\sum (x - \bar{x})^2}{N} = \frac{\sum x^2 - \frac{(\sum x)^2}{N}}{N} \tag{3}$$ Multiply $(3)$ through by $N$: $$\sum(x - \bar{x})^2 = \sum x^2 - \frac{(\sum x)^2}{N} \tag{4}$$ Expand the LHS in $(4)$: $$\sum\left(x^2 - 2x\bar{x} + \bar{x}^2\right) = {\sum x^2 - \frac{(\sum x)^2}{N}} \tag{5}$$ Expanding both sides in $(5)$: $$\sum x^2 - 2x\sum\bar{x} + \sum\bar{x}^2 = \sum x^2 - \frac{\sum x\sum x}{N} \tag{6}$$ From $(6)$: $$\sum\bar{x}^2 - 2\bar{x}\sum{x} = -\bar{x}\sum{x} \tag{7}$$ From $(7)$: $$\sum\bar{x}^2 = \bar{x}\sum{x} \tag{8}$$

I don't know how to make the LHS equal RHS in $(8)$.

Answer 1

$$\sum\bar{x}^2 = N \bar{x}^2$$

$$\bar{x}\sum x = \bar{x} \cdot N\bar{x} = N \bar{x}^2$$

Answer 2

You can also derive the second from the first: $$\sigma^2 = \frac{\sum (x - \bar{x})^2}{N}=\sigma^2 = \frac{\sum (x^2 - 2\cdot x\cdot\bar{x}+\bar{x}^2)}{N}=\frac{\sum x^2 -2\cdot\bar{x}\cdot \sum x+\sum \bar{x}^2}{N}=$$ $$\frac{\sum x^2 -2\cdot\frac{\sum x}{N}\cdot \sum x+ \bar{x}^2\cdot N}{N}=\frac{\sum x^2 -2\cdot\frac{(\sum x)^2}{N}+ \left(\frac{\sum x}{N}\right)^2\cdot N}{N}=$$ $$\frac{\sum x^2 -\frac{(\sum x)^2}{N}}{N}.$$