已知递增的等比数列{an}的前三项之积是64,且a2-1,a3-3,a4-9成等差数列. (1)求数列{an}的通项公式an; (2)设bn=n•an,求数列{bn}的前n项和Sn.

问题描述:

已知递增的等比数列{an}的前三项之积是64,且a2-1,a3-3,a4-9成等差数列.
(1)求数列{an}的通项公式an
(2)设bn=n•an,求数列{bn}的前n项和Sn

(1)设公比为q由题意得:a2=4,∵2(a3-3)=a2-1+a4-9,∴2(4q-3)=3+4q2-9,解得:q=2∴an=2n(2)∵Sn=b1+b2+…+bn=1×2+2×22+…+n×2n∴2Sn=1×22+2×23+…+(n-1)×2n+n×2n+1两式相减得,Sn=-2-22-23-…-2n...