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If 1×22+2×32+3×42+⋯+n(n+1)212×2+22×3+32×4+⋯+n2(n+1)=87, then n=

Answer» If 1×22+2×32+3×42++n(n+1)212×2+22×3+32×4++n2(n+1)=87, then n=


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