# Stuck on a term in $Var[\hat{\beta_o}]$ Proof

So I was trying to prove that $$Var[\hat{\beta_o}]=\dfrac{\sigma^2n^{-1}\sum{(x_i)^2}}{\sum{(x_i-\bar{x}})^2}$$

And I got stuck with the part $$\dfrac{ -2\bar{x}}{\sum{(x_i-\bar{x})^2}}\sum[{(x_i - \bar{x})}E(u_i\bar{u})]$$

Now, I need to show that this whole expression equals 0 but I don't know how to go about it. I'm mainly confused as to how I should be treating the $$E(u_i\bar{u})$$ term. I have read that $$u_i$$ and $$u_j$$ are uncorrelated so $$E(u_iu_j)=0$$ so long as $$i \ne j$$. But I don't know how I can use that fact to reduce the $$E(u_i\bar{u})$$ part to something. The $$u_i$$ I'm assuming would go on forever i.e. $$i=1,2,3........$$ but the $$\bar{u}$$ would only have the values of $$u$$ that go from $$1$$ to $$n$$ like $$u_1+u_2+....+u_n$$, so how should I go about this?

Note: $$u_i's$$ are NOT the residual terms. These are the actual $$u's$$ from the line/function that models the relationship between $$y$$ and $$x$$ And $$\bar{u}$$ would be defined to be actual $$u's$$ from $$1$$ to $$n$$ whole divided by $$n$$

Can anyone help simplify this? Thank you.

So at some point, you should get something like $$V(\hat{\beta_0}) = V(\bar{Y}) + \bar(x)^2V(\hat{\beta}_1) - 2\bar{x}\text{Cov}(\bar{Y}, \beta_1)$$.
I'm assuming that you are asking about the last term, but what you should get is $$\frac{\sigma^2}{n} \sum(\frac{x_i - \bar{x}}{S_{xx}})$$, which is slightly different than what you have, and it is the $$\sum({x_i - \bar{x}}) = 0$$ that causes the whole term to go to 0.