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,](/uploads/image/z/5430901-13-1.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97an%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2CSn%3Dn%5E2%2C%E6%B1%82%E5%92%8C1%2F%28a1a2%29%2B1%2F%28a2a3%29%2B.%2B1%2F%5B%28an-1an%5D+%28n%E2%89%A52+%29%E8%80%81%E5%B8%88%E8%AF%B4%2C%E7%94%A8%E8%A3%82%E9%A1%B9%E7%9B%B8%E6%B6%88%E6%B3%95%2C%E6%B1%82%E5%AE%8C%E6%95%B4%E8%BF%87%E7%A8%8B%2C)
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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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