已知数列{an}是等差数列,且a1=2,a1+a2+a3=12,令bn=2n•an,则数列{bn}的前n项和Sn=_.

问题描述:

已知数列{an}是等差数列,且a1=2,a1+a2+a3=12,令bn=2n•an,则数列{bn}的前n项和Sn=______.

∵数列{an}是等差数列,且a1=2,a1+a2+a3=12,∴2+2+d+2+2d=12,解得d=2,∴an=2+(n-1)×2=2n.∴bn=an•2n=n•2n+1,∴Sn=1×22+2×23+…+(n-1)×2n+n×2n+1,①2Sn=1×23+2×24+3×25+…+(n-1)×2n+1+n×2n+...