Summation Proof I'm getting stuck halfway through this:
Show that $$\sum_{i=1}^n (y_i - \bar y_s)^2 = \sum_{i=1}^n (y_i)^2 - n\bar y_s^2$$
My skills with manipulating sums are quite rusty. I multiply the left side and distribute the sum to each part. I can see that the middle term needs to become $-2n\bar y_s^2$ in order to combine with the third term to make $-n\bar y_s^2$, but I can't quite get there. 

I would say
$\displaystyle \sum_{i=1}^n (y_i - \bar y_s)^2 = \sum_{i=1}^n y_i^2-2\bar y_s\sum_{i=1}^ny_i+\sum_{i=1}^n \bar y_s^2= \sum_{i=1}^n y_i^2-2n\bar y_s^2+n\bar y_s^2=\sum_{i=1}^n y_i^2-n\bar y_s^2$
