已知{an}为等差数列,前n项和为Sn,a3=6,S3=12,an=2n(2)求1/S1+1/S2+…+1/Snan=2n 后面多了(2)
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![已知{an}为等差数列,前n项和为Sn,a3=6,S3=12,an=2n(2)求1/S1+1/S2+…+1/Snan=2n 后面多了(2)](/uploads/image/z/5463141-69-1.jpg?t=%E5%B7%B2%E7%9F%A5%7Ban%7D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2Ca3%3D6%2CS3%3D12%2Can%3D2n%282%29%E6%B1%821%2FS1%2B1%2FS2%2B%E2%80%A6%2B1%2FSnan%3D2n+%E5%90%8E%E9%9D%A2%E5%A4%9A%E4%BA%86%EF%BC%882%EF%BC%89)
已知{an}为等差数列,前n项和为Sn,a3=6,S3=12,an=2n(2)求1/S1+1/S2+…+1/Snan=2n 后面多了(2)
已知{an}为等差数列,前n项和为Sn,a3=6,S3=12,an=2n(2)求1/S1+1/S2+…+1/Sn
an=2n 后面多了(2)
已知{an}为等差数列,前n项和为Sn,a3=6,S3=12,an=2n(2)求1/S1+1/S2+…+1/Snan=2n 后面多了(2)
an=2n
Sn=2+4+6+……+2n=2*n(n-1)/2=n(n-1)
1/S1+1/S2+…+1/Sn=1/1*2+1/2*3+……+1/n(n-1)
=[1/1-1/2+1/2-1/3+1/3-1/4+……+1/(n-1)-1/n]
=1-1/n=(n-1)/n
已知an是等差数列,前n项和为Sn,求证:S3n=3(S2n-Sn)
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