Question

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 是什么意思 请讲解一下

Answer 1

an=n^2/(2n-1)(2n+1)
每个加号前叫一项
当n=1是an=1/1*3
当n=2是an=2平方/3*5
当n=3是an=3的平方/5*7
……
以此类推
本题n从1取到1004