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1212+(1+2)222+⋯+(1+2+3+⋯+16)2162= ? \dfrac{1^2}{1^2} + \dfrac{(1+2)^2}{2^2} + \cdots + \dfrac{(1+2+3+\cdots+16)^2}{16^2} = \, ? 1212+22(1+2)2+⋯+162(1+2+3+⋯+16)2=?
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