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$\sum y_k \sum \frac{x_j^2}{y_j} - \left(\sum x_k \right )^2 = \sum_{1\le i<j\le n}\frac{(x_i y_j-x_j y_i)^2}{y_i y_j}$
12-26-2010, 05:19 PM
Post: #1
$\sum y_k \sum \frac{x_j^2}{y_j} - \left(\sum x_k \right )^2 = \sum_{1\le i<j\le n}\frac{(x_i y_j-x_j y_i)^2}{y_i y_j}$
(1) $\displaystyle{\sum_{k=1}^n y_k \sum_{j=1}^n \frac{x_j^2}{y_j} - \left(\sum_{k=1}^n x_k \right )^2 = \sum_{1\le i<j\le n}\frac{(x_i y_j-x_j y_i)^2}{y_i y_j}}$
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