题目内容
已知无穷数列{an}的前n项和为Sn,且满足Sn=A
+Ban+C,其中A、B、C是常数.
(1)若A=0,B=3,C=-2,求数列{an}的通项公式;
(2)若A=1,B=
,C=
,且an>0,求数列{an}的前n项和Sn;
(3)试探究A、B、C满足什么条件时,数列{an}是公比不为-1的等比数列.
a | 2n |
(1)若A=0,B=3,C=-2,求数列{an}的通项公式;
(2)若A=1,B=
1 |
2 |
1 |
16 |
(3)试探究A、B、C满足什么条件时,数列{an}是公比不为-1的等比数列.
(1)∵Sn=A
+Ban+C,A=0,B=3,C=-2,
∴Sn=3an-2,
∴当n=1时,a1=3a1-2,解得a1=1;
当n≥2时,an=Sn-Sn-1=3an-3an-1,
整理,得2an=3an-1,
∴
=
,
∴an=(
)n-1.
(2)∵Sn=A
+Ban+C,A=1,B=
,C=
,
∴Sn=
+
an+
,
∴当n=1时,a1=
+
a1+
,解得a1=
,
当n≥2时,an=Sn-Sn-1=
-
+
an-
an-1
整理,得(an+an-1)(an-an-1-
)=0,
∵an>0,∴an-an-1=
,
∴{an}是首项为
,公差为
的等差数列,
∴Sn=
+
=
.
(3)若数列{an}是公比为q的等比数列,
①当q=1时,an=a1,Sn=na1
由Sn=A
+Ban+C,得na1=A
+Ba1+C恒成立
∴a1=0,与数列{an}是等比数列矛盾;
②当q≠±1,q≠0时,an=a1qn-1,Sn=
qn-
,
由Sn=A
+Ban+C恒成立,
得A×
×q2n+(B×
-
)×qn+C+
=0对于一切正整数n都成立
∴A=0,B=
≠1或
或0,C≠0,
事实上,当A=0,B≠1或
或0,C≠0时,
Sn=Ban+Ca1=
≠0,
n≥2时,an=Sn-Sn-1=Ban-Ban-1,
得
=
≠0或-1
∴数列{an}是以
为首项,以
为公比的等比数列.
a | 2n |
∴Sn=3an-2,
∴当n=1时,a1=3a1-2,解得a1=1;
当n≥2时,an=Sn-Sn-1=3an-3an-1,
整理,得2an=3an-1,
∴
an |
an-1 |
3 |
2 |
∴an=(
3 |
2 |
(2)∵Sn=A
a | 2n |
1 |
2 |
1 |
16 |
∴Sn=
a | 2n |
1 |
2 |
1 |
16 |
∴当n=1时,a1=
a | 21 |
1 |
2 |
1 |
16 |
1 |
4 |
当n≥2时,an=Sn-Sn-1=
a | 2n |
a | 2n-1 |
1 |
2 |
1 |
2 |
整理,得(an+an-1)(an-an-1-
1 |
2 |
∵an>0,∴an-an-1=
1 |
2 |
∴{an}是首项为
1 |
4 |
1 |
2 |
∴Sn=
n |
4 |
n(n-1) |
4 |
n2 |
4 |
(3)若数列{an}是公比为q的等比数列,
①当q=1时,an=a1,Sn=na1
由Sn=A
a | 2n |
a | 21 |
∴a1=0,与数列{an}是等比数列矛盾;
②当q≠±1,q≠0时,an=a1qn-1,Sn=
a1 |
q-1 |
a1 |
q-1 |
由Sn=A
a | 2n |
得A×
| ||
q2 |
a1 |
q |
a1 |
q-1 |
a1 |
q-1 |
∴A=0,B=
q |
q-1 |
1 |
2 |
事实上,当A=0,B≠1或
1 |
2 |
Sn=Ban+Ca1=
C |
1-B |
n≥2时,an=Sn-Sn-1=Ban-Ban-1,
得
an |
an-1 |
B |
B-1 |
∴数列{an}是以
C |
1-B |
B |
B-1 |
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