已知数列{an}时公差不为零的等差数列,a1=1,a1,a3,a9成等比数列,则数列{an•2an}的前n项和sn=_.
问题描述:
已知数列{an}时公差不为零的等差数列,a1=1,a1,a3,a9成等比数列,则数列{an•2an}的前n项和sn=______.
答
∵a1=1,a1,a3,a9成等比数列,∴a1a9=a23,即1+8d=(1+2d)2,∴4d=4d2,解得d=1,∴an=1+n-1=n,an•2an=n•2n,则sn=1⋅2+2⋅22+⋅⋅⋅+n⋅2n ①,2Sn=1⋅22+2⋅23+⋅⋅⋅+n⋅2n+1,②,两式相减得...