已知a=1,b=2,试求:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.1/(a+2008)(b+2008)的值.

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已知a=1,b=2,试求:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.1/(a+2008)(b+2008)的值.

已知a=1,b=2,试求:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.1/(a+2008)(b+2008)的值.
已知a=1,b=2,试求:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.1/(a+2008)(b+2008)的值.

已知a=1,b=2,试求:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.1/(a+2008)(b+2008)的值.
答案是2009/2010.
用错位相减法就行了,原式=1/2+(1/2-1/3)+(1/3-1/4)+...+(1/2009-1/2010).然后括号拆开就行了.


代入得
原式=1/1*2+1/2*3+1/3*4+...+1/2009*2010
=1-1/2010
=2009/2010