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. 
 A: Suppose that 
$$\bar y_s=\frac 1n\sum_{i=1}^{n}y_i.$$
Then, we have$$\begin{align}\sum_{i=1}^{n}(y_i-\bar y_s)^2&=\sum_{i=1}^{n}(y_i^2-2\bar y_sy_i+\bar y_s^2)\\&=\sum_{i=1}^{n}y_i^2-\sum_{i=1}^{n}2\bar y_sy_i+\sum_{i=1}^{n}\bar y_s^2\\&=\sum_{i=1}^{n}y_i^2-2\bar y_s\sum_{i=1}^{n}y_i+\bar y_s^2\sum_{i=1}^{n}1\\&=\sum_{i=1}^{n}y_i^2-2\bar y_s\cdot n\bar y_s+\bar y_s^2\cdot n\\&=\sum_{i=1}^{n}y_i^2-n\bar y_s^2.\end{align}$$
A: I would say
$\displaystyle \sum_{i=1}^n (y_i - \bar y_s)^2 = \sum_{i=1}^n y_i^2-2\bar y_s\sum_{i=1}^ny_i+\sum_{i=1}^n \bar y_s^2= \sum_{i=1}^n y_i^2-2n\bar y_s^2+n\bar y_s^2=\sum_{i=1}^n y_i^2-n\bar y_s^2$
A: Assuming that $\bar{y}_s$ is the average of all $y_i$, we can define 
$$a=\bar{y}_s=\frac1n\sum_{i=1}^ny_i$$
Hence, 
$$\begin{align}\sum_{i=1}^n(y_i-\bar{y}_s)^2&=\sum_{i=1}^n(y_i-a)^2\\
&=\sum_{i=1}^n(y_i^2-2ay_i+a^2)\\
&=\left(\sum_{i=1}^n y_i^2\right)-2a\underbrace{\left(\sum_{i=1}^n y_i\right)}_{na}+na^2\\
&=\left(\sum_{i=1}^n y_i^2\right)-2na^2+na^2\\
&=\left(\sum_{i=1}^n y_i^2\right)-na^2\\
&=\left(\sum_{i=1}^n y_i^2\right)-n\bar{y}_s^2\qquad \blacksquare\\
\end{align}$$
NB - Using $a$ to represent $\bar{y}_s$ simplifies the notation for the workings, such that $y$ only occurs in $y_i$ which is used only for summation, for ease of visual display. 
