(1)由条件知an=a1qn-1,0<q< ,a1>0,所以数列{an}是递减数列.若有ak,am,an(k<m<n)成等差数列,则中项不可能是ak(最大),也不可能是an(最小), 若2am=ak+an2qm-k=1+qn-k,(*) 由2qm-k≤2q<1,1+qh-k>1,知(*)式不成立, 故ak,am,an不可能成等差数列. (2)(ⅰ)(解法1)ak-ak+1-ak+2=a1qk-1(1-q-q2)=a1qk-1 , 由 ∈ ,知ak-ak+1-ak+2<ak<ak-1<…, 且ak-ak+1-ak+2>ak+2>ak+3>…, 所以ak-ak+1-ak+2=ak+1,即q2+2q-1=0, 所以q= -1. (解法2)设ak-ak+1-ak+2=am,则1-q-q2=qm-k, 由1-q-q2∈ 知m-k=1,即m=k+1, 以下同解法1. (ⅱ)bn= , (解法1)Sn=1+ + +…+ , Tn=1+ + +…+![](http://img.shitiku.com.cn/uploads/allimg/20191009/20191009023605-91782.png) =n+ =n -![](http://img.shitiku.com.cn/uploads/allimg/20191009/20191009023606-19911.png) =nSn-[(1- )+(1- )+(1- )+…+(1- )] =nSn- =nSn-![](http://img.shitiku.com.cn/uploads/allimg/20191009/20191009023606-70110.png) =nSn-n+Sn=(n+1)Sn-n,所以T2011=2012S2011-2011. (解法2)Sn+1=1+ =Sn+ ,所以(n+1)Sn+1-(n+1)Sn=1, 所以(n+1)Sn+1-nSn=Sn+1,2S2-S1=S1+1,3S3-2S2=S2+1,…… (n+1)Sn+1-nSn=Sn+1,累加得(n+1)Sn+1-S1=Tn+n, 所以Tn=(n+1)Sn+1-1-n=(n+1)Sn-n=(n+1)(Sn+bn)-1-n =(n+1) -1-n=(n+1)Sn-n, 所以T2011=2012S2011-2011 |