Question

若丨ab-2丨=(b-1）²=0，求ab分之1+（a+1）（b+1）分之1+（a+2）（b+2）分之+…+（a+2000）（b+2000)

分之1的值

Answer 1

丨ab-2丨+(b-1）²=0
ab-2=0,(b-1）²=0
b=1,
a-2=0
a=2

ab分之1+（a+1）（b+1）分之1+（a+2）（b+2）分之+…+（a+2000）（b+2000)
=1/1*2+1/2*3+1/3*4+..............+1/2001*2002
=1-1/2+1/2-1/3+1/3-1/4+..........+1/2001-1/2002
=1-1/2002
=2001/2002

Answer 2

丨ab-2丨=(b-1）²=0
所以b=1
a=2
1/ab+1/（a+1）（b+1）+1/（a+2）（b+2）+…+1/（a+2000）（b+2000)
=1/1*2+1/2*3+1/3*4+……+1/2001*2002
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/2001-1/2002)
=1-1/2002
=2001/2002

Answer 3

b=1，a=2.原式换为1/2+1/6+1/12+........=1-1/2+1/2-1/3+1/3-1/4+........+1/2001-1/2002=1-1/2002=2001/2002