已知数列{an}是等差数列,a3=5,a5=9.数列{bn}的前n项和为Sn,且Sn=1-bn2(n∈N*).(1)求数列{an}和{bn}的通项公式;(2)若cn=an•bn,求数列{cn}的前n项

题目简介

已知数列{an}是等差数列,a3=5,a5=9.数列{bn}的前n项和为Sn,且Sn=1-bn2(n∈N*).(1)求数列{an}和{bn}的通项公式;(2)若cn=an•bn,求数列{cn}的前n项

题目详情

已知数列{an}是等差数列,a3=5,a5=9.数列{bn}的前n项和为Sn,且Sn=
1-bn
2
(n∈N*)

(1)求数列{an}和{bn}的通项公式;
(2)若cn=an•bn,求数列{cn}的前n项和 Tn
题型:解答题难度:中档来源:汕头二模

答案

(1)法一:设数列的公差为d
由题意可得
a1++2d=5
a1+4d=9

解得a1=1,d=2
∴an=1+2(n-1)=2n-1
法二:设数列的公差是d
d=
a5-a3
5-3
=class="stub"9-5
2
=2

∴an=a5+2(n-5)=9+2n-10=2n-1
sn=
1-bn
2

当n=1时,b1=s1=
1-b1
2

∴b1=class="stub"1
3

当n≥2时,bn=sn-sn-1=class="stub"1
2
(1-bn)-class="stub"1
2
(1-bn-1)

=class="stub"1
2
(bn-1-bn)

bn
bn-1
=class="stub"1
3

∴数列{bn}是以class="stub"1
3
为首项,以class="stub"1
3
为公比的等比数列
∴bn=b1qn-1=(class="stub"1
3
)n

(2)cn=an•bn=class="stub"2n-1
3n

Tn=class="stub"1
3
+class="stub"3
32
+…+class="stub"2n-1
3n

class="stub"1
3
Tn
=class="stub"1
32
+class="stub"3
33
+…+class="stub"2n-3
3n
+class="stub"2n-1
3n+1
lll
两式相减可得,
2Tn
3
=class="stub"1
3
+2(class="stub"1
3
+class="stub"1
32
+…+class="stub"1
3n
)-class="stub"2n-1
3n+1

=class="stub"1
3
+
class="stub"2
9
(1-class="stub"1
3n-1
)
1-class="stub"1
3
-class="stub"2n-1
3n+1

=class="stub"2
3
-class="stub"2n+2
3n+1

Tn=1-class="stub"n+1
3n

更多内容推荐