已知等差数列an前n项和为Sn,Sn=n^2,求和1/(a1a2)+1/(a2a3)+.+1/[(an-1an] (n≥2 )老师_数学解答

已知等差数列an前n项和为Sn,Sn=n^2,求和1/(a1a2)+1/(a2a3)+.+1/[(an-1an] (n≥2 )
老师说,用裂项相消法,求完整过程,

人气：418 ℃　时间：2020-04-05 09:22:22

解答

n=1时,a1=S1=1²=1n≥2时,an=Sn-S(n-1)=n²-(n-1)²=2n-1n=1时,a1=2-1=1,同样满足通项公式数列{an}的通项公式为an=2n-11/[ana(n+1)]=1/[(2n-1)(2n+1)]=(1/2)[1/(2n-1)-1/(2n+1)]1/(a1a2)+1/(a2a3)+...+1...