Attempting to show $\frac 1N\sum_{n=1}^N x_n^2 - \overline x^2 = \frac 1N\sum_{n=1}^N (x_n-\overline x)^2$

Question

I'm reading through this pdf on the least squares method and I don't see how they've gotten one step in their derivation.

At the end of page 5 they've written

$$\frac 1N\sum_{n=1}^N x_n^2 - \overline x^2 = \frac 1N\sum_{n=1}^N (x_n-\overline x)^2$$

But here's what I get when I try to verify this:

$$\frac 1N\sum x_n^2 - \overline x^2 = \frac 1N\sum x_n^2 - \overline x^2\left(\frac 1N \sum 1\right) = \frac 1N\sum (x_n^2-\overline x^2)$$

But $$\frac 1N\sum (x_n-\overline x)^2 = \frac 1N\sum (x_n^2-2x_n\overline x+ \overline x^2) \\ = \frac 1N\left[\sum x_n^2-2\overline x\left(\frac{N}{N}\sum x_n\right)+ \sum\overline x^2\right] = \frac 1N\left[\sum x_n^2-2N\sum \overline x^2+ \sum\overline x^2\right] \\ = \frac 1N\left[\sum x_n^2-(2N-1)\sum \overline x^2\right]$$

That last part doesn't look like it would equal $\frac 1N\sum(x_n^2-\overline x^2)$ unless $N=1$. Where am I going wrong?

Answer 1

Hint: $$2\overline x \left(\frac{N}{N} \sum x_n\right)\neq 2N\sum \overline x^2$$

Answer 2

$$ x_{n}^{2} = \left[\left(x_{n} - \overline{x}\right) + \overline{x}\right]^{2} = \left(x_{n} - \overline{x}\right)^{2} + 2\left(x_{n} - \overline{x}\right)\overline{x} + \overline{x}^{2} = \left(x_{n} - \overline{x}\right)^{2} + 2x_{n}\,\overline{x} - \overline{x}^{2} $$ Sum over $n$ in both sides and divide by $N$.