考察一般项：(3n-2)[3(n+1)-2][3(n+2)-2]=(3n-2)(3n+1)(3n+4)=(9n²-3n-2)(3n+4)=27n³+27n²-18n-81×4×7+4×7×10+...+22×25×28=27×(1³+2³+...+8³)+27×(1²+2²+...+8
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