因为数列{an}是等差数列 2a2=a1+a3
a1+a2+a3=12 3a2=12 a2=4
an=2+(n-1)*2=2n
bn=2nx^n
1）若x=1 则bn=2n sn=2^(n+1)-2
2)若x≠1
sn=2x+4x²+6x³+……+2（n-1)x^(n-1)+2nx^n
xsn=2x²+4x³+……+2(n-1)x^n+2nx^(n+1)
sn-xsn=2x+2x²+2x³+……+2x^n-2nx^(n+1)=[2x(1-x^n)/(1-x)]-2nx^(n+1)
两边同时除以1-x 得sn=[2x(1-x^n)/(1-x)²]-2nx^(n+1)/(1-x)