sigma squared notation i wanna to know From where came No. 2 in this:
$$E\left( \sum (x_i - \bar x) \right)^2 = E\left(\sum(x_i - \bar x)^2 +2\sum\sum (x_i - \bar x)(x_j - \bar x)\right)$$
and why in this removed 
$$2\sum\sum\rho \sigma^2)= n(n-1)\rho\sigma^2$$
 A: I would recommend first working through simple summations with $ n = 2 $, $ n = 3 $, etc. in order to identify patterns in these simple cases.
n = 2
$ E \big( \sum_{i=1}^{2} (x_i-\bar{x}) \big)^2 = E \big( (x_1-\bar{x}) + (x_2-\bar{x}) \big)^2 = E \big( \big[ (x_1-\bar{x}) + (x_2-\bar{x}) \big] \big[ (x_1-\bar{x}) + (x_2-\bar{x}) \big] \big) = E \big( (x_1-\bar{x})^2 + (x_1-\bar{x})(x_2-\bar{x}) + (x_2-\bar{x})(x_1-\bar{x}) + (x_2-\bar{x})^2 \big) = E \big((x_1-\bar{x})^2 + (x_2-\bar{x})^2 + 2(x_1-\bar{x})(x_2-\bar{x}) \big) $
The above line can be re-written as:
$$ E \big( \sum_{i=1}^{2} (x_i - \bar{x})^2 + 2 \underset{i\neq j}{\sum\sum} (x_i-\bar{x})(x_j-\bar{x}) \big) $$
n = 3
$ E \big( \sum_{i=1}^{3} (x_i-\bar{x}) \big)^2 = E \big( (x_1-\bar{x}) + (x_2-\bar{x}) + (x_3-\bar{x}) \big)^2 = E \big( \big[ (x_1-\bar{x}) + (x_2-\bar{x}) + (x_3-\bar{x}) \big] \big[ (x_1-\bar{x}) + (x_2-\bar{x}) + (x_3-\bar{x}) \big] \big) = E \big( (x_1-\bar{x})^2 + (x_1-\bar{x})(x_2-\bar{x}) + (x_1-\bar{x})(x_3-\bar{x}) + (x_2-\bar{x})(x_1-\bar{x}) + (x_2-\bar{x})^2 + (x_2-\bar{x})(x_3-\bar{x}) + (x_3-\bar{x})(x_1-\bar{x}) + (x_3-\bar{x})(x_2-\bar{x}) + (x_3-\bar{x})^2 \big) = E \big((x_1-\bar{x})^2 + (x_2-\bar{x})^2 + (x_3-\bar{x})^2 + 2(x_1-\bar{x})(x_2-\bar{x}) + 2(x_1-\bar{x})(x_3-\bar{x}) + 2(x_2-\bar{x})(x_3-\bar{x}) \big) $
The above line can be re-written as:
$$ E \big( \sum_{i=1}^{3} (x_i - \bar{x})^2 + 2 \underset{i\neq j}{\sum\sum} (x_i-\bar{x})(x_j-\bar{x}) \big) $$
