比较上下两行算式,你发现了什么中
1/2-1/3=(3-2)/(2*3)=1/(2*3),1/3-1/4=(4-3)/(3*4)=1/(3*4),1/4-1/5=(5-4)/(4&5)=1/(4*5)
规律1/n-1/(n+1)=[(n+1)-n]/[n(n+1)]=1/[n(n+1)]
由此可见
原式=(2-1)/(1*2)+(3-2)/(2*3)+(4-3)/(3*4)+(5-4)/(4*5)+……|(100-99)/(99*100)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)++……+(1/99-1/100)
=1-1/100
=99/100。 全部
1/2-1/3=(3-2)/(2*3)=1/(2*3),1/3-1/4=(4-3)/(3*4)=1/(3*4),1/4-1/5=(5-4)/(4&5)=1/(4*5)
规律1/n-1/(n+1)=[(n+1)-n]/[n(n+1)]=1/[n(n+1)]
由此可见
原式=(2-1)/(1*2)+(3-2)/(2*3)+(4-3)/(3*4)+(5-4)/(4*5)+……|(100-99)/(99*100)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)++……+(1/99-1/100)
=1-1/100
=99/100。
收起