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OpenStudy (anonymous):
Prove that [\sum_{i}^{}m_ir_i]^2=(\sum_{i}^{}m_i)(\sum_{i}^{}m_ir_i^2)-\frac{1}{2}\sum_{i,j}^{}m_im_jr_{ij}^2
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OpenStudy (anonymous):
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OpenStudy (anonymous):
whats that
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