m-1的绝对值+n-2的平方=0
m=1,n=2,
mn分之1+(m+1)(n+1)分之1+…+(m+2009)(n+2009)分之1
=1/(1*2)+1/[(1+1)(2+1)]+……+1/[(1+2009)(2+2009)]
=1/2+(1/2-1/3)+(1/3-1/4)+……+(1/2008-1/2009)
=1/2+1/2-1/3+1/3-1/4+……+1/2007-1/2008+1/2008-1/2009
=1-1/2009
=2008/2009
m-1的绝对值+n-2的平方=0
m=1,n=2,
mn分之1+(m+1)(n+1)分之1+…+(m+2009)(n+2009)分之1
=1/(1*2)+1/[(1+1)(2+1)]+……+1/[(1+2009)(2+2009)]
=1/2+(1/2-1/3)+(1/3-1/4)+……+(1/2008-1/2009)
=1/2+1/2-1/3+1/3-1/4+……+1/2007-1/2008+1/2008-1/2009
=1-1/2009
=2008/2009