(n+1)b
(Ⅰ)因为Sn=2n+1-2, 所以,当n=1时,a1=S1=21+1-2=2=21, 当n≥2时, an=Sn-Sn-1=2n+1-2n=2n,(2分) 又a1=S1=21+1-2=2=21,也满足上式, 所以数列{an}的通项公式为an=2n.(3分) b1=a1=2,设公差为d,则由b1,b3,b9成等比数列, 得(2+2d)2=2×(2+8d),(4分) 解得d=0(舍去)或d=2,(5分) 所以数列{bn}的通项公式为bn=2n.(6分) (Ⅱ)cn= 2 (n+1)bn= 1 n(n+1)(8分) 数列{cn}的前n项和: Tn= 1 1×2+ 1 2×3+ 1 3×4+…+ 1 n×(n+1)(10分) =1- 1 2+ 1 2− 1 3+…+ 1 n− 1 n+1=1- 1 n+1= n n+1.(12分)
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