The Curie-Weiss model from Ising

Question

Can anyone explain why this is the case?

$$ \begin{aligned} p(\mathbf{x}) &=\frac{\exp \left\{\sum_{i=1}^{n} \mu_{i} x_{i}+\sum_{i=1}^{n} \sum_{j=1}^{n} \sigma x_{i} x_{j}\right\}}{\sum_{\mathbf{x}} \exp \left\{\sum_{i=1}^{n} \mu_{i} x_{i}+\sum_{i=1}^{n} \sum_{j=1}^{n} \sigma x_{i} x_{j}\right\}} \\ &=\frac{\exp \left\{\sum_{i=1}^{n} \mu_{i} x_{i}+\sigma x_{+}^{2}\right\}}{\sum_{\mathbf{x}} \exp \left\{\sum_{i=1}^{n} \mu_{i} x_{i}+\sigma x_{+}^{2}\right\}}, \end{aligned} $$ in which $x_+ = \sum_{i = 1}^{n} x_i$ refers to the sum of the node states

Answer 1

because $$ \begin{aligned} \sum_{i=1}^{n} \sum_{j=1}^{n} x_{i} x_{j} &= x_1^2 + x_1x_2 + x_1x_3 + \dots + x_1x_n \\ &+ x_1x_2 +x_2^2+x_2x_3+ \dots x_2x_n\\ & + \dots + \\ & x_1x_n+x_2x_n+ x_3x_n+ \dots + x_n^2 \\ &= (x_1+x_2 + \dots + x_n)(x_1+x_2 + \dots + x_n)\\ &=(x_1+x_2 + \dots + x_n)^2. \end{aligned} $$