摘要：已知数列{an}满足:a1＝.a2＝.an+1＝2an-an-1(n≥2.n∈N*).数列{bn}满足b1＜0,3bn-bn-1＝n(n≥2.n∈N*).数列{bn}的前n项和为Sn. (1)求数列{an}的通项an, (2)求证:数列{bn-an}为等比数列. 解:(1)证明∵2an＝an+1+an-1(n≥2.n∈N*). ∴{an}是等差数列. 又∵a1＝.a2＝.∴an＝+(n-1)·＝. (2)证明:∵bn＝bn-1+(n≥2.n∈N*). ∴bn+1-an+1＝bn+-＝bn- ＝(bn-)＝(bn-an). 又∵b1-a1＝b1-≠0. ∴{bn-an}是以b1-为首项.以为公比的等比数列.

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