请问[1/(2+4)]+[1/(4+6)]+[1/(6+8)]+…+[1/(198+200)]=?[1/(2+4)]+[1/(4+6)]+[1/(6+8)]+…+[1/(198+200)]=多少啊?

来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/19 23:35:14
请问[1/(2+4)]+[1/(4+6)]+[1/(6+8)]+…+[1/(198+200)]=?[1/(2+4)]+[1/(4+6)]+[1/(6+8)]+…+[1/(198+200)]=多少啊?请

请问[1/(2+4)]+[1/(4+6)]+[1/(6+8)]+…+[1/(198+200)]=?[1/(2+4)]+[1/(4+6)]+[1/(6+8)]+…+[1/(198+200)]=多少啊?
请问[1/(2+4)]+[1/(4+6)]+[1/(6+8)]+…+[1/(198+200)]=?
[1/(2+4)]+[1/(4+6)]+[1/(6+8)]+…+[1/(198+200)]=多少啊?

请问[1/(2+4)]+[1/(4+6)]+[1/(6+8)]+…+[1/(198+200)]=?[1/(2+4)]+[1/(4+6)]+[1/(6+8)]+…+[1/(198+200)]=多少啊?
一楼的答案是错的,但我也想不出一个好方法.

原式=1/2*[2/(2+4]+[2/(4+6)]+[2/(6+8)]+…+[2/(198+200)]
=1/2*[1/4-1/6+1/6-1/8+1/8-1/10+1/10-……-1/200]
=1/2*[1/4-1/200]
=49/400