(1-1/2+1/3-1/4+...+1/99-1/100)/(1/101^2-1^2+1/102^2-2^2+...+
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/08/15 06:47:05
(1-1/2+1/3-1/4+...+1/99-1/100)/(1/101^2-1^2+1/102^2-2^2+...+1/150^2-50^2)
![(1-1/2+1/3-1/4+...+1/99-1/100)/(1/101^2-1^2+1/102^2-2^2+...+](/uploads/image/z/4689099-27-9.jpg?t=%281-1%2F2%2B1%2F3-1%2F4%2B...%2B1%2F99-1%2F100%29%2F%281%2F101%5E2-1%5E2%2B1%2F102%5E2-2%5E2%2B...%2B)
分子
=(1+1/2+1/3+1/4+……+1/99+1/100)-2*(1/2+1/4+……+1/100)
=1+1/2+1/3+..+1/100-(1/1+1/2+1/3+...+1/50)
=1/51+1/52+…+1/99+1/100
分母
=1/(100-1)(101+1)+1/(102-2)(102+2)+...+1/(150-50)(150+50)
=1/100(1/102+1/104+...+1/200)
=1/200(1/51+1/52+...+1/100)
所以,
原式=(1/51+1/52+…+1/99+1/100)/[1/200(1/51+1/52+...+1/100)]
=1/(1/200)
=200
=(1+1/2+1/3+1/4+……+1/99+1/100)-2*(1/2+1/4+……+1/100)
=1+1/2+1/3+..+1/100-(1/1+1/2+1/3+...+1/50)
=1/51+1/52+…+1/99+1/100
分母
=1/(100-1)(101+1)+1/(102-2)(102+2)+...+1/(150-50)(150+50)
=1/100(1/102+1/104+...+1/200)
=1/200(1/51+1/52+...+1/100)
所以,
原式=(1/51+1/52+…+1/99+1/100)/[1/200(1/51+1/52+...+1/100)]
=1/(1/200)
=200
(1-1/2+1/3-1/4+...+1/99-1/100)/(1/101^2-1^2+1/102^2-2^2+...+
1-1/2+1/3-1/4+1/5-1/6+…+1/99-1/100 1/(101+1)+1/(2+102)+1/(3+
(-1)+2+(-3)+4+(-5)+6+.+(-99)+100+(-101)+102
(1-1/2+1/3-1/4+1/5-...+1/99-1/100)/[1/(1+101)+1/(2+102)+1/(3
1-1/2+1/3-1/4+...+1/99-1/100/1/1+101+1/2+102+1/3+103+...+1/5
(1/1+101+1/2+102+……+1/50+150)÷(1-1/2+1/3-1/4+……+1/99-1/100)
1,2,3,...,99,100; 2,3,4,...,100,101; 4,5,6,...,101,102;...10
(1+1/1*3)*(1+1/2*4)*(1+1/3*5).(1+1/98*100)*(1+1/99*101)
(1+1/1+3)*(1+1/2*4)*(1+1/3*5)*.*(1+1/98*100)*(1+1/99*101)
1/1*2*3+1/2*3*4+1/3*4*5+...+1/99*100*101
1/1*2*3+1/2*3*4+1/3*4*5+.1/99*100*101
1 2 3 .99 100 2 3 4 .100 101 3 4 5 .101 102 .100 101 102 .19