问题
解答题
在等比数列{an}中,a2a3=32,a5=32.
(1)求数列{an}的通项公式;
(2)设数列{an}的前n项和为Sn,求S1+2S2+…+nSn.
答案
(1)设等比数列{an}的公比为q,依题意
,解得a1=2,q=2,a1q•a1q2=32 a1q4=32
∴an=2•2n-1=2n.
(2)∵a1=2,q=2,
∴Sn=
=2(2n-1),2(1-2n) 1-2
∴S1+2S2+…+nSn=2[(2+2•22+…+n•2n)-(1+2+…+n)],
设Tn=2+2•22+…+n•2n,①
则2Tn=22+2•23+…+n•2n+1,②
①-②,得-Tn=2+22+…+2n-n•2n+1=
-n•2n+1=(1-n)•2n+1-2,2(1-2n) 1-2
∴Tn=(n-1)•2n+1+2,
∴S1+2S2+…+nSn=2[(n-1)•2n+1+2]-n(n+1)=(n-1)•2n+2+4-n(n+1).