1/1*3+2平方/3*5+3的平方/5*7+……+1004的平方/2007*2009
考察一般项：
an=n^2/(2n-1)(2n+1)
=n^2/(4n^2-1)
=(1/4)(4n^2-1+1)/(4n^2-1)
=(1/4)[1+1/(4n^2-1)]
=(1/4)+(1/4)*[1/(2n-1)(2n+1)]
=(1/4)+(1/8)[1/(2n-1)-1/(2n+1)]
对于本题,n=1004
1/1*3+2^2/3*5+3^2/5*7+……+1004^2/2007*2009
=1004*(1/4)+(1/8)(1-1/3+1/3-1/5+1/5-1/7+...+1/2005-1/2007+1/2007-1/2009)
=251+(1/8)(1-1/2009)
=251+251/2009
=251(1+1/2009)
=251*2010/2009
=504510/2009
为什么用1004*(1/4)+(1/8)(1-1/3+1/3-1/5+1/5-1/7+...+1/2005-1/2007+1/2007-1/2009)