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
问题描述：
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
答
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.