提供一个通用方法吧:
考察一般项第k项:k(k+1)(k+2)=k³+3k²+2k
1×2×3+2×3×4+...+n(n+1)(n+2)
=(1³+2³+...+n³)+3(1²+2²+...+n²)+2(1+2+...+n)
=[n(n+1)/2]²+3n(n+1)(2n+1)/6+2n(n+1)/2
=n²(n+1)²/4 +n(n+1)(2n+1)/2 +n(n+1)
=[n(n+1)/4][n(n+1)+2(2n+1)+4]
=[n(n+1)/4](n²+5n+6)
=n(n+1)(n+2)(n+3)/4 以上推导可以用于任意多项相加
对于本题:
1×2×3+2×3×4+...+7×8×9=(7×8×9×10)/4=1260
解:因为1x2x3=(1x2x3×4-0x1x2×3)/4
2x3x4=(2x3x4×5-1x2x3×4)/4
.........
7x8x9=(7x8x9×10-6x7x8x9)/4
所以 1x2x3+2x3x4+3x4x5+…+7x8x9
=(1x2x3×4-0x1x2×3)/4+(2x3x4×5-1x2x3×4)/4+...(7x8x9×10-6x7x8x9)/4
=(7x8x9×10)/4
=1260
1x2x3+2x3x4+3x4x5+…+7x8x9
=(1x2x3×4-0x1x2×3)/4+(2x3x4×5-1x2x3×4)/4+...(7x8x9×10-6x7x8x9)/4
=(7x8x9×10)/4
=1260