limx3-1 x2 x 3

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

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

解1/(1+2+3)+1/(2+3+4)+1/(3+4+5)……+1/(99+100+101)=1/6+1/9+1/12+1/15.+1/300=1/3*(1/2+1/3.+1/99+1/100)=1

解题思路：本题主要依靠分式，将分母进行拆分，然后逐一消除，得到简便运算解题过程：1/1*2+2/1*2*3+3/1*2*3*4+……+6/1*2*…*10=1-1/2+(1/1*2-1/2*3)+(1

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

1×2×3+2×3×4+3×4×5+……+8×9×10=(1/4)(1×2×3×4)+(1/4)(2×3×4×5-1×2×3×4)+(1/4)(3×4×5×6-2×3×4×5)+……(1/4)(8×9

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

1X2+2X3+3X4+、、、、、、+nX(n+1）=(1/3)(1*2*3-0*1*2)+(1/3)(2*3*4-1*2*3)+(1/3)(3*4*5-2*3*4)+.+(1/3)[n*(n+1)(

最简方法:拆项法n(n+1)(n+2)=1/4*[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]...2x3x4=1/4*[2x3x4*5-1*2x3x4]1x2x3=1/4*[

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

设第n项为anan=n(n+1)(n+2)=n^3+3n^2+2n1×2×3+2×3×4+...+10×11×12=(1^3+2^3+...+10^3)+3(1^2+2^2+...+10^2)+2(1

=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

乘积因子中只要包含2和5,则乘积的个位数字一定是0,因此1×2×3×...×100,1×2×3×...9,……,1×2×3×4×5个位数字均为0,需要考察的仅仅是1×2×3×4,1×2×3,1×2,1

（1）1x2+2x3+…+99x100+100x101==1/3x100x101x102=343400（2）1x2+2x3+3x4+…+n(n+1)(n为正整数)=1/3n(n+1)(n+2)（3）1

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

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