(Ⅰ)在an+1=2an+1两边同时加上1,构造出an+1+1=2(an+1),易证明数列{an+1}为等比数列.
(Ⅱ)结合分组法求和及错位相消法求和计算.
(Ⅲ)由已知可得出bn+2-2bn+1+bn=0,继而{bn}是等差数列,通项公式易求.
【解析】
(Ⅰ)∵an+1=2an+1(n∈N*).an+1+1=2(an+1),----------(3分)
{an+1}是以a1+1=2为首项,2为公比的等比数列.∴.
即.--------------(4分)
(II)∵,cn=2n,∴
∴Sn=a1c1+a2c2+a3c3+…+ancn=2[(1×2+2×22+3×23+…+n×2n)-(1+2+3+…+n)]-----(6分)
设 A=1×2+2×22+3×23+…+n×2n①
则2A=1×22+2×23+…+(n-1)×2n+n×2n+1②
①-②得-A=1×2+1×22+1×23+…+1×2n-n×2n+1==(1-n)×2n+1-2
∴A=(n-1)×2n+1+2
∴--------------(9分)
(Ⅲ)∵,∴,
∴2[(b1+b2+…+bn)-n]=nbn,①2[(b1+b2+…+bn+bn+1)-(n+1)]=(n+1)bn+1. ②
②-①,得2(bn+1-1)=(n+1)bn+1-nbn,--------------(11分)
即(n-1)bn+1-nbn+2=0,③nbn+2-(n+1)bn+1+2=0. ④
④-③,得nbn+2-2nbn+1+nbn=0,
即bn+2-2bn+1+bn=0,∴,∴{bn}是等差数列.--------------(13分)
∵b1=2,b2=4,∴bn=2n.--------------(15分)
(注:没有证明数列{bn}是等差数列,直接写出bn=2n,给2分)