Are you aware that the sum of deviations from the mean is always zero? (That's something I do to check the correctness of my work when I am finding standard deviation by hand.) If not, it is easy to prove by breaking up [MATH]\sum_{i=1}^n(x_i-\bar{x})[/MATH] as [MATH]\sum_{i=1}^n x_i-\sum_{i=1}^n\bar{x}[/MATH].
As a result, [MATH]\overline{Y}[/MATH] is being multiplied by zero.