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,

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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,
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,

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,
n=1时,a1=S1=1²=1
n≥2时,an=Sn-S(n-1)=n²-(n-1)²=2n-1
n=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-1
1/[ana(n+1)]=1/[(2n-1)(2n+1)]=(1/2)[1/(2n-1)-1/(2n+1)]
1/(a1a2)+1/(a2a3)+...+1/[a(n-1)an]
=(1/2)[1/1-1/3+1/3-1/5+...+1/(2n-3)-1/(2n-1)] 这步就是裂项了.
=(1/2)[1 -1/(2n-1)]
=(n-1)/(2n-1)

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