$$ \left(\sum^n_{i=1}{a_i}^2\right)\cdot\left(\sum^n_{i=1}{b_i}^2\right)-\left(\sum^n_{i=1}{a_i}{b_i}\right)^2=\sum_{i
An identity involving products of sums
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$\begingroup$
algebra-precalculus
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0Try induction in $n$. – 2012-12-08