Showing $s^2=\frac{1}{n-1}\sum_{i=1}^n(x_i-\bar{x})^2=\frac{1}{n-1}\left [\sum_{i=1}^n x_i^2-\frac{1}{n}\left ( \sum_1^nx_i\right )^2 \right ]$

Question

I've got as far as this $$s^2=\frac{1}{n-1}\sum_{i=1}^n(x_i-\bar{x})^2=\frac{1}{n-1}\sum_{i=1}^n\left ( x_i^2-2x_i\bar{x}+\bar{x}^2\right )$$ $$=\frac{1}{n-1}\left [\sum_{i=1}^nx_i^2+\bar{x}^2-2\bar{x}\sum_{i=1}^n x_i \right ]$$ $$\frac{1}{n-1}\left [\sum_{i=1}^nx_i^2 +\left ( \bar{x}-\sum_{i=1}^n x_i\right )^2 -\left (\sum_{i=1}^n x_i \right )^2\right ]$$ And now I'm stuck. I can use $$\sum_{i=1}^n\frac{x_i}{n}=\bar{x}$$ However can't get the result. If someone could point me in the right direction I'd be grateful.

Answer 1

Your second line should be

$$=\frac{1}{n-1}\left [\sum_{i=1}^nx_i^2+n\bar{x}^2-2\bar{x}\sum_{i=1}^n x_i \right ]$$

Then just as you said, you can use

$$\sum_{i=1}^n\frac{x_i}{n}=\bar{x}$$

to continue.