设数列{an}的前N项和为Sn,已知1/Sn+1/S2+1/S3+.+1/Sn=n/(n+1),求Sn

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设数列{an}的前N项和为Sn,已知1/Sn+1/S2+1/S3+.+1/Sn=n/(n+1),求Sn设数列{an}的前N项和为Sn,已知1/Sn+1/S2+1/S3+.+1/Sn=n/(n+1),求

设数列{an}的前N项和为Sn,已知1/Sn+1/S2+1/S3+.+1/Sn=n/(n+1),求Sn
设数列{an}的前N项和为Sn,已知1/Sn+1/S2+1/S3+.+1/Sn=n/(n+1),求Sn

设数列{an}的前N项和为Sn,已知1/Sn+1/S2+1/S3+.+1/Sn=n/(n+1),求Sn
由1/S1+1/S2+1/S3+.+1/Sn=n/(n+1),知,当n=1时,s1=2,
当n≥2时1/S1+1/S2+1/S3+.+1/Sn-1=(n-1)/n,两式相减得,1/sn =1/[n(n+1)]
所以sn=n(n+1).,此式也适合s1.

1/Sn = (1/S1+1/S2+1/S3+......+1/Sn ) - (1/Sn+1/S2+1/S3+......+1/Sn-1) = n/(n+1) - (n-1)/n
即 1/Sn= 1/n(n+1)
所以Sn = n(n+1)
如 S1=2 , S2 = 6
呵呵,解数学题有时只要换个思路就好了

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