已知數(shù)列﹛an﹜的前n項和為Sn,a1=1,且2nSn+1－2(n+1)Sn=n²+n(n∈N*).（1）求數(shù)列{an}的通項公式；（2）設(shè)bn=n/2(n+3)Sn,求數(shù)列{bn}的前n項和Tn；（3）證明：n≥2時,1/（a2

已知數(shù)列﹛an﹜的前n項和為Sn,a1=1,且2nSn+1－2(n+1)Sn=n²+n(n∈N*).（1）求數(shù)列{an}的通項公式；（2）設(shè)bn=n/2(n+3)Sn,求數(shù)列{bn}的前n項和Tn；（3）證明：n≥2時,1/（a2的三次方）+1/（a3的三次方）+1/(a4的三次方）+……+1/（an的三次方）＜1/4

數(shù)學(xué)人氣：340 ℃時間：2019-08-19 17:08:58

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