(Ⅰ)由题意可得,a1=a2,a1+a2=a3
(Ⅱ)由Sn=an+1=Sn+1-Sn,可得2Sn=Sn+1,=2,从而可得{Sn}为等比数列,进而可求
(Ⅲ)由(II)可得,Sn=(2n-1)=2n-2,bn=n-2,从而可求cn=+n2n-2,令A=++…+,利用分组求和,令B=1•2-1+2•2+3•21+4•22+…+n2n-2,利用错位相减可求,从而可求
【解析】
(Ⅰ)∵a1=a2,a1+a2=a3,∴2a1=a3=1,∴a1=,a2=.…(4分)
(Ⅱ)∵Sn=an+1=Sn+1-Sn,∴2Sn=Sn+1,=2,…(6分)
∴{Sn}是首项为,公比为2的等比数列.
∴Sn=2n-1=2n-2.…(8分)
(Ⅲ)Sn=(2n-1)=2n-2,bn=n-2,bn+3=n+1,bn+4=n+2,
∵cn•bn+3•bn+4=1+n(n+1)(n+2)Sn,∴cn•(n+1)(n+2)=1+n(n+1)(n+2)2n-2,
即cn=+n2n-2.…(10分)
令A=++…+=-++…+
=-.…(12分)
令B=1•2-1+2•2+3•21+4•22+…+n2n-2,①
2B=1•2+2•21+3•22+…+(n-1)2n-2+n2n-1,②
②-①得
B=n2n-1-2-1-2-21-…-2n-2=n2n-1-=(n-1)2n-1+,
∴c1+c2+…+cn=-+(n-1)2n-1+=(n-1)2n-1+.…(14分)