已知數列{an}的前n項和爲Sn,a1=2,Sn=n2+n.(Ⅰ)求數列{an}的通項公式;(Ⅱ)設的前n項和...
2021-08-30
問題詳情:已知數列{an}的前n項和爲Sn,a1=2,Sn=n2+n.(Ⅰ)求數列{an}的通項公式;(Ⅱ)設的前n項和爲Tn,求*Tn<1.【回答】解:(Ⅰ)∵Sn=n2+n,∴當n≥2時,an=Sn-Sn-1=n2+n-(n-1)2-(n-1)=2n,又a1=2滿足上式,∴an=2n(n∈N*).(Ⅱ)*:∵Sn=n2+n=n(n+1),∴=-,∴Tn=++…+=1-.∵...