已知数列｛An}为等差数列,且A1=2,A1+A2+A3=12,令Bn=An·3的n次方（n属于整数）求｛Bn｝的前n项和的公式.

a(n) = 2 + (n-1)d.
s(n) = 2n + n(n-1)d/2.
12 = s(3) = 6 + 3d,d=2.
a(n)=2 + 2(n-1) = 2n.
b(n) = a(n)3^n = 2n*3^n,
t(n) = b(1)+b(2)+b(3)+...+b(n-1)+b(n)
=2*1*3 + 2*2*3^2 + 2*3*3^3 + ...+ 2(n-1)3^(n-1) + 2n3^n,
3t(n) = 2*1*3^2 + 2*2*3^3 + ...+ 2(n-1)3^n + 2n3^(n+1).
2t(n) = 3t(n) - t(n) = -2*1*3 - 2*1*3^2 - 2*1*3^3 - ...- 2*1*3^n + 2n3^(n+1)
= 2n3^(n+1) - 6[1+3+...+3^(n-1)]
= 2n3^(n+1) - 6[3^n - 1]/(3-1)
= 2n3^(n+1) - 3[3^n - 1]
= 3 + (2n-1)3^(n+1),
t(n) = 3/2 + [(2n-1)/2]3^(n+1)