# Show explicitly that the following identity holds under a Simple Linear Regression: sum_(i=1)^n r_i mu_i=0 with residuals r_i=y_i-mu_i and mu_i=beta_0+beta_1 x_i

Show explicitly that the following identity holds under a Simple Linear Regression:

with residuals ${r}_{i}={y}_{i}-\stackrel{^}{{\mu }_{i}}$ and $\stackrel{^}{{\mu }_{i}}=\stackrel{^}{{\beta }_{0}}+\stackrel{^}{{\beta }_{1}}{x}_{i}$.
my steps:

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After the third equality, you should use the linearity of the sum and notice that ${\stackrel{^}{\beta }}_{0}$ and ${\stackrel{^}{\beta }}_{1}$ don't depend on the index i. That way, you should get:
$\sum _{i=1}^{n}{r}_{i}{\stackrel{^}{\mu }}_{i}={\stackrel{^}{\beta }}_{0}n\overline{y}+{\stackrel{^}{\beta }}_{1}\sum _{i=1}^{n}{x}_{i}{y}_{i}-n{\stackrel{^}{\beta }}_{0}^{2}-2{\stackrel{^}{\beta }}_{0}{\stackrel{^}{\beta }}_{1}n\overline{x}-{\stackrel{^}{\beta }}_{1}^{2}\sum _{i=1}^{n}{x}_{i}^{2}.$
Now, you should use the fact that ${\stackrel{^}{\beta }}_{0}=\overline{y}-{\stackrel{^}{\beta }}_{1}\overline{x}$ and remember the normal equations, namely the second one:
${\stackrel{^}{\beta }}_{0}n\overline{x}+{\stackrel{^}{\beta }}_{1}\sum _{i=1}^{n}{x}_{i}^{2}-\sum _{i=1}^{n}{x}_{i}{y}_{i}=0.$
The result now directly follows.