已知数列{an}中,a1=1,a2=3且2an+1=an+2+an(n∈N+)数列{bn}的前n项和为Sn,其中b1=-32,bn+1=-23Sn(n∈N+).(1)求数列{an}和{bn}的通项公式

题目简介

已知数列{an}中,a1=1,a2=3且2an+1=an+2+an(n∈N+)数列{bn}的前n项和为Sn,其中b1=-32,bn+1=-23Sn(n∈N+).(1)求数列{an}和{bn}的通项公式

题目详情

已知数列{an}中,a1=1,a2=3且2an+1=an+2+an(n∈N+)数列{bn}的前n项和为Sn,其中b1=-
3
2
bn+1=-
2
3
Sn(n∈N+).

(1)求数列{an}和{bn}的通项公式;
(2)若Tn=
a1
b1
+
a2
b2
+…+
an
bn
,求Tn
的表达式.
题型:解答题难度:中档来源:不详

答案

(1)∵2an+1=an+2+an∴数列{an}是等差数列,(1分)
∴公差d=a2-a1=2∴an=2n-1 (3分)
∵bn+1=-class="stub"2
3
Sn∴bn=-class="stub"2
3
Sn-1(n≥2)
bn+1-bn=-class="stub"2
3
bn,∴bn+1= class="stub"1
3
bn

又∵b2=-class="stub"2
3
S1=1
b2
b1
=-class="stub"2
3
≠class="stub"1
3

∴数列{bn}从第二项开始是等比数列,
bn=
-class="stub"3
2
,(n=1)
(class="stub"1
3
)n-2,(n≥2)
(6分)
(2)∵n≥2时
an
bn
=(2n-1)•3n-2
(7分)∴Tn=
a1
b1
+
a2
b2
++
an
bn
=-class="stub"2
3
+3×30+5×31+7×32++(2n-1)×3n-2

∴3Tn=-2+3×31+5×32+7×33++(2n-1)×3n-1(10分)
错位相减并整理得Tn=-class="stub"2
3
+(n-1)×3n-1
.(12分)

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