1乘4分之5加4乘7分之29加7乘10分之71加10乘13分之137

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/

=[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)-

5/7×4+5/7+2×7/5=5/7×（1+4+2）=5/7×7=5

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/

=1-1/2+1/2-1/3+1/3-1/4+……+1/9-1/10=1-1/10=9/10

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

1x1/4=1/3(1-1/4)1/4x1/7=1/3(1/4-1/7)...照此规律,原式可化为：1/3(1-1/4)+1/3(1/4-1/7)+...+1/3(1/91-1/94)=1/3(1-1

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+……

乘以3分之4减去3乘以5分之8加上5乘以7分之12减去7乘以9分之16=4/3-(4/3-4/5)+(6/5-6/7)-(8/7-8/9)=4/5+6/5-(6/7+8/7)+8/9=2-2+8/9=

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))

3乘4分之1乘6分之1=3/4x1/6=1/84分之3乘8分之5乘32=15/32x32=15（7分之1加5分之4加35分之2）乘9分之7=1/7x7/9+4/5x7/9+2/35x7/9=1/9+2

2+5\84

7分之4乘5加7分之2乘6加7分之1乘24=7分之1乘20加7分之1乘12加7分之1乘24=7分之1×(20+12+24)=7分之1×56=8

原式=1/2-1/3+1/3-1/4+1/4-1/5+...+1/2010-1/2011+1/2011-1/2012=1/2-1/2012=1005/2012

1/(1x4)+1/(4x7)+1/(7x10)+...+1/(2005x2008)=1/3*(1/1-1/4)+1/3*(1/4-1/7)+1/3*(1/7-1/10)+...+1/3*(1/200

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

(5分之4加10分之9)乘20=（4/5+9/10)*20=4/5*20+9/10*20=16+18=34（2分之1+7分之1）乘2乘7=1/2*2*7+1/7*2*7=7+2=9