x 3 x 4=364

n(n+1)(n+2)(n+3)/4

任何三个相领自然数的积必定是6的陪数

原式=﹙1-1／2-1／3﹚+﹙1／2-1／3-1／4﹚+﹙1／3-1／4-1／5﹚+······+﹙1／11-1／12-13／13﹚=1-1／13=12／13

一般的,有：(n-1)n(n+1)=n^3-n{n^3}求和公式：Sn=[n(n+1)/2]^2{n}求和公式：Sn=n(n+1)/21x2x3+2x3x4+3x4x5+.+7x8x9=2^3-2+3

2和3因为三个连续自然数至少有一个是偶数,且大于或等于2,所以它们的乘积一定是2的倍数三个连续自然数有一个是3的倍数,所以它们的乘积一定是3的倍数

1/n(n+1)(n+2)=1/2*[1/n-2/(n+1)+1/(n+2)]原式=1/2*(1-2*1/2+1/3+1/2-2*1/3+1/4+.+1/9-2*1/10+1/11)=1/2*(1-1

nx(n+1)=1/3[n(n+1)(n+2)-(n-1)n(n+1)]1x2+2x3+3x4+...+nx(n+1)=1/3[1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+

3*(1x2+2x3+3x4+...+99x100)=3*1/3*(1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+99x100x101-98x99x100)=99x100x1

因为1x2x3=(1x2x3×4-0x1x2×3)/42x3x4=(2x3x4×5-1x2x3×4)/4.7x8x9=(7x8x9×10-6x7x8x9)/4所以1x2x3+2x3x4+3x4x5+…

1x2x3/1x3x4=2x4x6/2x6x8=------=nx2nx3n/nx3nx4n=1/2[(1x2x3+2x4x6+...+nx2nx3n)/(1x3x4+2x6x8+...+nx3nx4

原式=1/4（-0*1*2*3+1*2*3*4）+1/4（-1*2*3*4+2*3*4*5）+……+1/4[-（n-1)n(n+1)(n+2)+n(n+1)(n+2)(n+3)]=1/4[-0*1*2

裂项求和1/((n-1)n(n+1))=(1/2)*(1/((n-1)n)-1/(n(n+1)))接着便可算了结果我算的是(1/2)*(1/6-1/(49*50))

（4）+（6）=3*4*5+5*6*7=60+210=270

6/7

=1/2×(1/1×2-1/2×3+1/2×3-1/3×4+1/3×4-1/4×5+1/4×5-1/5×6)=1/2×(1/1×2-1/5×6)=7/30

1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+.1/(20*21*22)=1/2[1/1*2-1/2*3]+1/2[1/2*3-1/3*4]+1/2[1/3*4-1/4*5]+...+1

1/1×2×3=1/2×(1/1×2-1/2×3)其他以此类推所以原式=1/2×(1/1×2-1/2×3+1/2×3-1/3×4+……+1/20×21-1/21×22)=1/2×(1/1×2-1/21

1/1x2x3+1/2x3x4+1/3x4x5+.1/98x99x100==1/2[1/1*2-1/2*3]+1/2[1/2*3-1/3*4]+1/2[1/3*4-1/4*5]+...+1/2[1/9

1/1x2x3+1/2x3x4+1/3x4x5+1/4x5x6+.+1/48x49x50=48*51÷(4*49*50)=306/1225

=1/2×[1/1×2-1/2×3+1/2×3-1/3×4+……+1/n(n+1)-1/(n+1)(n+2)]=1/2×[1/1×2-1/(n+1)(n+2)]=1/4-2/(4n²+12n