(Ⅰ)设数列{a n}公差为d,
则a 1+a 2+a 3=3a 1+3d=12,
又a 1=2,d=2,
∴a n=2n,
(Ⅱ)由(1)可得 b n = a n + 2 n =2n+2 n,
∴S n=2(1+2+…+n)+(2+2 2+…+2 n)=n(n+1)+2 n+1-2=2 n+1+n 2+n-2.
(Ⅰ)设数列{a n}公差为d,
则a 1+a 2+a 3=3a 1+3d=12,
又a 1=2,d=2,
∴a n=2n,
(Ⅱ)由(1)可得 b n = a n + 2 n =2n+2 n,
∴S n=2(1+2+…+n)+(2+2 2+…+2 n)=n(n+1)+2 n+1-2=2 n+1+n 2+n-2.