Question

若丨a-1丨+丨ab-2丨=0求ab分之1+（a+1)(b+1)分之1+（a+2）（b+2）+···+（a+2002）（b+2002）分之1的值

Answer 1

丨a-1丨+丨ab-2丨=0
两个非负数之和为0，则每个都为0
a-1=0
ab-2=0
a=1,b=2
ab分之1+（a+1)(b+1)分之1+（a+2）（b+2）+···+（a+2002）（b+2002）分之1
=1/1*2+1/2*3+1/3*4+-----+1/2003*2004
=1-1/2+1/2-1/3+1/3-1/4+------+1/2003-1/2004
=1-1/2004
=2003/2004

Answer 2

首先，a=1,b=2
即问题等价于求(*是×)
1/(1*2)+1/(2*3)+1/(3*4)+···+1/(2003*2004)
其中1/(n*(n+1))=(1/n)-(1/n+1)
故1/(1*2)+1/(2*3)+1/(3*4)+···+1/(2003*2004)=(1-1/2)+(1/2-1/3)+(1/3-1/4))+···+(1/2003-1/2004)
=1-1/2004=2003/2004
这是裂项相消
采纳采纳

Answer 3

由丨a-1丨+丨ab-2丨=0可知a=1,b=2. ab分之1+（a+1)(b+1)分之1+（a+2）（b+2）+···+（a+2002）（b+2002）分之1=1/(1*2)+1/(2*3)+1/(3*4)+...+1/(2003*2004)=(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+...(1/2003-1/2004)=1/1-1/2004=2003/2004(方法：数列求和中的裂项相消法）