解:(1)证明:由Sn+2-(t+1)Sn+1+tSn=0,得tSn+1-tSn=Sn+2-Sn+1,即an+2=tan+1,而a1=t,a2=t2, ∴数列{an}是以t为首项,t为公比的等比数列, ∴an=tn. (2)∵(tn+t-n)-(2n+2-n)=(tn-2n)[1-( )n],又 <t<2, <1, 则tn-2n<0且1-( )n>0, ∴(tn-2n)[1-( )n]<0, ∴tn+t-n<2n+2-n. (3)证明:∵ = (tn+t-n), ∴2( + +…+ )<(2+22+…2n)+(2-1+2-2+…+2-n)=2(2n-1)+1-2-n =2n+1-(1+2-n)<2n+1-2 , ∴ + +…+ <2n-![](http://img.shitiku.com.cn/uploads/allimg/20191016/20191016045019-41274.png) |