设正项等比数列{an}的首项a1=12,前n项和为Sn,且-a2,a3,a1成等差数列.(Ⅰ)求数列{an}的通项;(Ⅱ)求数列{nSn}的前n项和Tn.-数学

题目简介

设正项等比数列{an}的首项a1=12,前n项和为Sn,且-a2,a3,a1成等差数列.(Ⅰ)求数列{an}的通项;(Ⅱ)求数列{nSn}的前n项和Tn.-数学

题目详情

设正项等比数列{an}的首项a1=
1
2
,前n项和为Sn,且-a2,a3,a1成等差数列.
(Ⅰ)求数列{an}的通项;
(Ⅱ)求数列{nSn}的前n项和Tn
题型:解答题难度:中档来源:浙江二模

答案

(Ⅰ)设设正项等比数列{an}的公比为q(q>0),由题有a1-a2=2a3,且a1=class="stub"1
2

a1-a1q=2a1q2,即有2q2+q-1=0,解得q=-1(舍去)或q=class="stub"1
2

an=class="stub"1
2n

(Ⅱ)因为是首项、公比都为class="stub"1
2
的等比数列,故Sn=
class="stub"1
2
(1-class="stub"1
2n
)
1-class="stub"1
2
=1-class="stub"1
2n
,nSn=n-class="stub"n
2n

则数列{nSn}的前n项和 Tn=(1+2+…+n)-(class="stub"1
2
+class="stub"2
22
+…+class="stub"n
2n
)

Tn
2
=class="stub"1
2
(1+2+…+n)-(class="stub"1
22
+class="stub"2
23
+…+class="stub"n-1
2n
+class="stub"n
2n+1
)

前两式相减,得  
Tn
2
=class="stub"1
2
(1+2+…+n)-(class="stub"1
2
+class="stub"1
22
+…+class="stub"1
2n
)+class="stub"n
2n+1
=
n(n+1)
4
-
class="stub"1
2
(1-class="stub"1
2n
)
1-class="stub"1
2
+class="stub"n
2n+1

Tn=
n(n+1)
2
+class="stub"1
2n-1
+class="stub"n
2n
-2

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