己知|ab-2|+(b-2)的平方=0,求:1/ab+1/(a+1)*(b+1)+1/(a+2)*(b+2)+……+1/(a+2003)*(b+2003)的值

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己知|ab-2|+(b-2)的平方=0,求:1/ab+1/(a+1)*(b+1)+1/(a+2)*(b+2)+……+1/(a+2003)*(b+2003)的值
己知|ab-2|+(b-2)的平方=0,求:1/ab+1/(a+1)*(b+1)+1/(a+2)*(b+2)+……+1/(a+2003)*(b+2003)的值

己知|ab-2|+(b-2)的平方=0,求:1/ab+1/(a+1)*(b+1)+1/(a+2)*(b+2)+……+1/(a+2003)*(b+2003)的值
由题意知:
ab -2 =0;
b -2 =0.
由此解得:a =1, b =2.
因为: 1/(1 +n)*(2 +n) = 1/(1 +n) -1/(2 +n).
所以
原式=1/1*2 + 1/2*3 + 1/3*4 + . + 1/2004*2005
=1 -1/2 + 1/2 - 1/3 + ... + 1/2004 -1/2005
=1 -1/2005
=2004/2005.