(Ⅰ)分别令n=2,n=3,n=4即可求得;
(Ⅱ)由an+1=2an+3,得an+1+3=2(an+3),根据等比数列的定义可作出证明;
(Ⅲ)由(Ⅱ)知,an+3=5•2n-1,变形可得an;
(Ⅰ)【解析】
由an+1=2an+3得,a2=2a1+3=7,a3=2a2+3=17,a4=2a3+3=37;
(Ⅱ)由an+1=2an+3,得an+1+3=2(an+3),
又a1+3=5,知=2,
所以数列{an+3}是以5为首项,2为公比的等比数列.
(Ⅲ)【解析】
由(Ⅱ)知,an+3=5•2n-1,
所以an=5•2n-1-3;