5分之4乘7.7加3.3乘.08减5分之4

1/(1*2*3)+1/(2*3*4)an=1/[n(n+1)(n+2)]=[(n+2)-(n+1)]/[n(n+1)(n+2)]=1/n(n+1)-1/n(n+2)=[1/n-1/(n+1)]-(1

1乘3乘5分之4加3乘5乘7分之4+...+95乘97乘99分之4=1/(1×3)-1/(3×5)+1/(3×5)-1/(5×7)+...+1/(95×97)-1/(97×99)=1/(1×3)-1/

7.7乘5分之1加0.43乘22=7.7×0.2+0.43×22=1.1×7×0.2+0.43×22=22×(0.7+0.43)=22×0.5=11

=[1/(1*3)-1/(3*5)]+[1/(3*5)-1/(5*7)]+.+[1/(95*97)-1/(97*99)]=1/(1*3)-[1/(3*5)-1/(3*5)]-.-[1/(95*97)-

原式=(4/5)×7.7+3.3×(0.8)-4/5=(4/5)×(7.7+3.3-1)=(4/5)×10=8

(1*2*3+2*4*6+3*6*9+4*8*12)/(3*4*5+6*8*10+9*12*15+12*16*20)=(1*2*3)(1+8+27+64)/(3*4*5)(1+8+27+64)=(1*

5/7×4+5/7+2×7/5=5/7×（1+4+2）=5/7×7=5

1,=(1/2)×(1/1×2+1/2×3)+(1/2)×(1/2×3-1/3×4)+……+(1/2)×(1/48×49-1/49×50)=(1/2)×(1/1×2+1/2×3+1/2×3-1/3×4

3乘4乘5分之一加4乘5乘6分之1加5乘6乘7分之1加6乘7乘8分之1.98乘99乘100分之1=1/2[1/3*4-1/4*5]+1/2[1/4*5-1/5*6]+...+1/2[1/98*99-1

利用公式：2/n(n+1)(n+2)=1/n(n+1)-1/(n+1)(n+2)1/(4×5×6)+1/(5×6×7)+…+1/(98×99×100)=1/2[1/4*5-1/5*6+1/5*6-1/

1乘2分之1加2乘3分之1加3乘4分之1加4乘5分之1加5乘6分之1加6乘7分之1=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7=1-1/7=6/7=7分

原式=1/2*(1/2*3-1/3*4+1/3*4-1/4*5+.+1/8*9-1/9*10)=1/2*(1/6-1/90)=1/2*(14/90)=7/90

=1-2分之1+2分之1-3分之1+3分之1-4分之1+……+99分之1-100分之1=1-100分之1=100分之99

1/2*3+1/3*4+1/4*5+……+1/49*50=(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+……+(1/49-1/50)=1/2-1/3+1/3-1/4+1/4-1/5+……

0.91乘99加91乘0.01=0.91*（99+1）=0.91*100=911.25乘88加12除以5分之4=1.25*88+12*1.25=1.25*（88+12）=1.25*100=125

4949/19800因为1/n(n+1)(n+2)=1/n(1/(n+1)-1/(n+2))=1/n-1/(n+1)-1/2(1/n-1(n+1))

2+5\84

2/(1×2×3)+2/(2×3×4)+...+2/(28×29×30)=2*1/2*[1/(1×2)-1/(2×3)+1/(2×3)-1/(3×4)+...+1/(28×29)-1/(29×30)]

看到你问过一个类似的题目,三个连续自然数相乘的倒数,此题类似,考虑通式：（2n+3)/[n(n+1)(n+2)(n+3)]=(n+n+3)/[n(n+1)(n+2)(n+3)]=1/[(n+1)(n+

=1*2*3(1+2*2*2+7*7*7)/[1*3*5(1+2*2*2+7*7*7)]=1*2*3/(1*3*5)=2/5