已知/a-1/+/b-2/=0,试求1/ab+1/(a+1)(b+1)1/(a+2)(b+2)+...+1/(a+2010)(b+2010)的值

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已知/a-1/+/b-2/=0,试求1/ab+1/(a+1)(b+1)1/(a+2)(b+2)+...+1/(a+2010)(b+2010)的值

已知/a-1/+/b-2/=0,试求1/ab+1/(a+1)(b+1)1/(a+2)(b+2)+...+1/(a+2010)(b+2010)的值
已知/a-1/+/b-2/=0,试求1/ab+1/(a+1)(b+1)1/(a+2)(b+2)+...+1/(a+2010)(b+2010)的值

已知/a-1/+/b-2/=0,试求1/ab+1/(a+1)(b+1)1/(a+2)(b+2)+...+1/(a+2010)(b+2010)的值
1/ab+1/(a+1)(b+1)1/(a+2)(b+2)+...+1/(a+2010)(b+2010)?
应该是:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2010)(b+2010)吧
/a-1/+/b-2/=0 可得a=1...b=2 b-a=1
1/(a+n)(b+n)=1/(a+n)-1/(b+n) b+n=a+n+1
所以原式=1/a-1/b+1/(a+1)-1/(b+1)+.+1/(a+2010)-1/(b+2010)
=1/a-1/b+1/b-1/(b+1)+...+1/(b+2009)-1/(b+2010)
=1/a-1/(b+2010)
=1-1/2012=2011/2012