(Ⅰ)由an+1=4an-3n+1可得an+1-(n+1)=4an-3n+1-(n+1)=4an-4n=4(an-n),从而可证
(Ⅱ)由(Ⅰ)可求an,利用分组求和及等差数列与等比数列的求和公式可求Sn,进而可证不等式.
(Ⅰ)证明:∵an+1=4an-3n+1n∈N*,
∴an+1-(n+1)=4an-3n+1-(n+1)=4an-4n=4(an-n)
∴{an-n}为首项a1-1=1,公比q=4的等比数列;
(Ⅱ)【解析】
∵an-n=4n-1,∴an=n+4n-1,
∴Sn=1+2+…+n+(1+4+…+4n-1)=+
∴Sn+1-4Sn=+-4[+]=-+2≤0
∴Sn+1≤4Sn.