[基本原理:其實所有數(shù)列求和的方法基本都一樣:裂項求和.這個數(shù)列也不例外.所謂裂項求和,就是先裂項,再求和.求和過程中必會相互消去,從而簡化.] 解: (裂項)由已知易得: an=1/（2n-1）(2n+1) =0.5*[(2n+1)-(2n-1)]/(2n+1)(2n-1) =0.5*[(2n+1)/(2n+1)(2n-1)-(2n-1)/(2n+1)(2n-1)] =0.5*[1/(2n-1)-1/(2n+1)] ∴a1=0.5*(1-1/3) a2=0.5*(1/3-1/5) a3=0.5*(1/5-1/7) a4=0.5*(1/7-1/9) …… an=0.5*[1/(2n-1)-1/(2n+1)] ∴Sn=a1+a2+a3+……+an =0.5*[(1-1/3)+(1/3-1/5)+(1/5-1/7)+……+1/(2n-1)-1/(2n+1)] =0.5*[1-1/3+1/3-1/5+1/5-1/7+……+1/(2n-1)-1/(2n+1)] =0.5*[1-1/(2n+1)] =n/(2n+1) 即Sn=n/(2n+1)
采納哦