如果有理数a,b满足|ab-2|+|1-b|=0.试求1/ab+1/(a+1)(b+1)+1(a+2)(b+2)+…+1/(a+2008)(b+2008)的值就是这个答案|ab-2|+|1-b|=0.1-b=0,b=1ab-2=0,ab=2,a=21/ab+1/(a+1)(b+1)+1(a+2)(b+2)+…+1/(a+2008)(b+2008)=1-1/2+1/2-1/3+.+1/2009-1/2010=1-1/201

Question

如果有理数a,b满足|ab-2|+|1-b|=0.试求1/ab+1/(a+1)(b+1)+1(a+2)(b+2)+…+1/(a+2008)(b+2008)的值
就是这个答案|ab-2|+|1-b|=0.
1-b=0,b=1
ab-2=0,ab=2,a=2
1/ab+1/(a+1)(b+1)+1(a+2)(b+2)+…+1/(a+2008)(b+2008)
=1-1/2+1/2-1/3+.+1/2009-1/2010
=1-1/2010
=2009/2010
我有些看不懂,求ab的值是明白了
但是那个规律没看懂,为什么就是=1-1/2+1/2-1/3+.+1/2009-1/2010,从哪里得出来的,规律是什么啊,

Answer 1

a=2,b=1
然后我们看规律
1/ab+1/(a+1)(b+1)+1(a+2)(b+2)
先看着3个
1/(1*2) +1/(3*2)+1/(4*3)
1/(1*2) 对应 1-(1/2)
1/(3*2) 对应 1-(5/6)=(1/2)-(1/3)
1/(4*3) 对应 1-(11/12)=(1/3)-(1/4)
1/(4*3) 对应 1-(11/12)=(1/3)-(1/4)
可明白了吗