Show that the standard formula of the OLS estimator of the slope can be rewritten as: $$ \frac{\sum_{i=1}^{n}(X_i - \bar{X})Y_i}{\sum_{i=1}^{n}(X_i-\bar{X})^2} $$ after using (and proving) that $\sum_{i=1}^{n}(X_i-\bar{X})Z = 0$ for a non-varying variable $Z$.
My work below, starting with the proof.
\begin{align*} & \sum_{i=1}^{n}(X_i-\bar{X})Z = 0\\ & \sum_{i=1}^{n}X_iZ-\sum_{i=1}^{n}\bar{X}Z = 0\\ & \sum_{i=1}^{n}X_iZ = n\bar{X}Z\\ & \frac{1}{n}\sum_{i=1}^{n}X_iZ = \bar{X}Z\\ & \bar{X}Q = \bar{X}Q\\ \end{align*}
Now that that's out of the way...
\begin{align*} \beta_1 & = \frac{\sum_{i=1}^{n}(X_i-\bar{X})(Y_i-\bar{Y})}{\sum_{i=1}^{n}(X_i-\bar{X})^2}\\ & = \frac{\sum_{i=1}^{n}X_iY_i-\bar{Y}\sum_{i=1}^{n}X_i - \bar{X}\sum_{i=1}^{n}Y_i + n\bar{X}\bar{Y}}{\sum_{i=1}^{n}X_i^2 - 2\bar{X}\sum_{i=1}^{n}X_i + n\bar{X}^2}\\ & = \frac{\frac{1}{n}\sum_{i=1}^{n}X_iY_i-\frac{\bar{Y}}{n}\sum_{i=1}^{n}X_i - \frac{\bar{X}}{n}\sum_{i=1}^{n}Y_i + \bar{X}\bar{Y}}{\frac{1}{n}\sum_{i=1}^{n}X_i^2 - 2\frac{\bar{X}}{n}\sum_{i=1}^{n}X_i + \bar{X}^2}\\ & = \frac{\frac{1}{n}\sum_{i=1}^{n}X_iY_i - \bar{X}\bar{Y}}{\frac{1}{n}\sum_{i=1}^{n}X_i^2 - 2\bar{X}^2}\\ & = \frac{\sum_{i=1}^{n}X_iY_i - n\bar{X}\bar{Y}}{\sum_{i=1}^{n}X_i^2 - 2n\bar{X}^2}\\ & = \frac{\sum_{i=1}^{n}X_iY_i - \frac{1}{n}\sum_{i=1}^{n}X_i\sum_{i=1}^{n}Y_i}{\sum_{i=1}^{n}X_i^2 - \frac{1}{n}(\sum_{i=1}^{n}X_i)^2}\\ &* = \frac{\sum_{i=1}^{n}X_iY_i - \bar{X}\sum_{i=1}^{n}Y_i}{\sum_{i=1}^{n}X_i^2 - \frac{1}{n}(\sum_{i=1}^{n}X_i)^2}\\ & = \frac{\sum_{i=1}^{n}(X_i - \bar{X})Y_i}{\sum_{i=1}^{n}X_i^2 - \frac{1}{n}(\sum_{i=1}^{n}X_i)^2}\\ & = \frac{\sum_{i=1}^{n}(X_i - \bar{X})Y_i}{\sum_{i=1}^{n}X_i^2 - 2n\bar{X}^2}\\ & = \frac{\sum_{i=1}^{n}(X_i - \bar{X})Y_i}{\frac{1}{n}\sum_{i=1}^{n}X_i^2 - 2\bar{X}^2}\\ & = \frac{\sum_{i=1}^{n}(X_i - \bar{X})Y_i}{\frac{1}{n}\sum_{i=1}^{n}X_i^2 - 2\frac{\bar{X}}{n}\sum_{i=1}^{n}X_i + \bar{X}^2}\\ & = \frac{\sum_{i=1}^{n}(X_i - \bar{X})Y_i}{\sum_{i=1}^{n}(X_i-\bar{X})^2}\\ \end{align*}
I don't see how the proof was useful to the problem. I feel that I may have messed up when pulling out the $Y_i$ variable from the sum because it is varying, unlike $Z$ (the line is starred). Would appreciate any feedback.
It can be much shorter. Recall that $\sum_{i=1}^nx_i=n\bar{x}_n$, thus
\begin{align} \hat{\beta}_1=&\frac{\sum(x_i-\bar{x})(Y_i-\bar{Y})}{\sum(x_i-\bar{x})^2}\\ =&\frac{\sum (x_i-\bar{x})Y_i - \sum (x_i-\bar{x})\bar{Y}}{\sum(x_i-\bar{x})^2}\\ =&\frac{\sum (x_i-\bar{x})Y_i - n\bar{x}\bar{Y}+n\bar{x}\bar{Y}}{\sum(x_i-\bar{x})^2}\\ =&\frac{\sum(x_i-\bar{x})Y_i}{\sum(x_i-\bar{x})^2}. \end{align}