因为60等于3x4x5,

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

1\1x2x3=2x(1\1x2-1\2x3),所以：1\1x2x3+1\2x3x4+1\3x4x5+``````+1\48x49x50=2x(1\1x2-1\2x3)+2x(1\2x3-1\3x4)

∵1/（2×3×4）=0.5×（1/2-2/3+1/4）1/（3×4×5）=0.5×(1/3-2/4+1/5)1/（4×5×6）=0.5×(1/4-2/5+1/6)…………∴上式=0.5×（1/2-2

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

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

1/2x3X4=（1/2X3-1/3X4）X1/21/3X4x5=（1/3X4-1/4X5）X1/2.1/8X9X10=（1/8X9-1/9X10）X1/2上述各式相加得：1/2x3X4十1/3X4x

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

1/1x2x3+1/2x3x4+1/3x4x5+1/4x5x6+...+1/8x9x10=1-1/2-1/2(1-1/3)+1/2-1/3-1/2(1/2-1/4)+1/3-1/4-1/2(1/3-1

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))

=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

c语言main(){inta,b,c,max,sum;sum=0;a=1;b=2;c=3;scanf("%d",&max);for(;max

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

错,4不是质数