Minimization of the SSE

Question

I'm reading the demonstration in the Book of Wackerly about the minimization of the $\mathrm{SSE}$ for the least square method but there's is a step I don't understand why an $n$ appears where I highlighted after the differentiation. I'm completely lost. Thanks for your help

\begin{align*} \frac{\partial\mathrm{SSE}}{\partial\hat\beta_0} & =\frac{\partial\left\{\sum_{i=1}^n[y_i-(\hat\beta_0+\hat\beta_1x_i)]^2\right\}}{\partial\hat\beta_0}=-\sum_{i=1}^n2[y_i-(\hat\beta_0+\hat\beta_1x_i)]\\ & =-2\left(\sum_{i=1}^ny_i-\color{red}{n\hat\beta_0}-\hat\beta_1\sum_{i=1}^nx_i\right)=0 \end{align*} and \begin{align*} \frac{\partial\mathrm{SSE}}{\partial\hat\beta_1} & =\frac{\partial\left\{\sum_{i=1}^n[y_i-(\hat\beta_0+\hat\beta_1x_i)]^2\right\}}{\partial\hat\beta_1}=-\sum_{i=1}^n2[y_i-(\hat\beta_0+\hat\beta_1x_i)]x_i\\ & =-2\left(\sum_{i=1}^nx_iy_i-\hat\beta_0\sum_{i=1}^nx_i-\hat\beta_1\sum_{i=1}^nx_i^2\right)=0. \end{align*}

Answer 1

$$\begin{split}-2\sum (y-(\beta_0+\beta_1 x))&=-2\left(\sum y-\beta_0\sum -\beta_1\sum x\right)\text{ distribute}\\ &=-2\left(\sum y-\left(\underbrace{\beta_0+\beta_0+...+\beta_0}_{n}\right)-\beta_1\sum x\right)\\ &=-2\left(\sum y-n\beta_0-\beta_1\sum x\right)\end{split}$$

Answer 2

$$-\sum_{i=1}^n 2[y_i - (\hat{\beta_0} + \hat{\beta_1} x_i)]$$ $$= -2[(y_1 - (\hat{\beta_0} + \hat{\beta_1} x_1)) + (y_2 - (\hat{\beta_0} + \hat{\beta_1} x_2)) + \ldots + (y_n - (\hat{\beta_0} + \hat{\beta_1} x_n))]$$

Can you see the reason now?