求23题详细解答 20
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2x3+3x12+4x27+......+(n+1)x3n²
=3+3+12+2x12+27+3x27+......+3n²+nx3n²
=3+12+27+......+3n²+3+2x12+3x27+......+nx3n²
=(3+12+27+......+3n²)+(3+2x12+3x27+......+nx3n²)
=3(1+4+9+......+n²)+3(1+2x4+3x9+......+nxn²)
=3(1²+2²+3²+......+n²)+3(1³+2³+3³+......+n³)
=3n(n+1)(2n+1)/6+3n²(n+1)²/4
=n(n+1)(2n+1)/2+3n²(n+1)²/4
=2n(n+1)(2n+1)/4+3n²(n+1)²/4
=n(n+1)[2(2n+1)+3n(n+1)]/4
=n(n+1)[4n+2+3n²+3n]/4
=n(n+1)(3n²+7n+2)/4
=n(n+1)(n+2)(3n+1)/4
=3+3+12+2x12+27+3x27+......+3n²+nx3n²
=3+12+27+......+3n²+3+2x12+3x27+......+nx3n²
=(3+12+27+......+3n²)+(3+2x12+3x27+......+nx3n²)
=3(1+4+9+......+n²)+3(1+2x4+3x9+......+nxn²)
=3(1²+2²+3²+......+n²)+3(1³+2³+3³+......+n³)
=3n(n+1)(2n+1)/6+3n²(n+1)²/4
=n(n+1)(2n+1)/2+3n²(n+1)²/4
=2n(n+1)(2n+1)/4+3n²(n+1)²/4
=n(n+1)[2(2n+1)+3n(n+1)]/4
=n(n+1)[4n+2+3n²+3n]/4
=n(n+1)(3n²+7n+2)/4
=n(n+1)(n+2)(3n+1)/4
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