判断1.因为60 =3x4x5,所以3,4,5都

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

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

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+

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

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

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

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))

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

错,4不是质数

答案：1.4949/99002.43.24.45.571.由1/n(n+1)(n+2)=[1/n-2/(n+1)+1/(n+2)]/2,得1/1x2x3+1/2x3x4+1/3x4x5+...+1/9