题目内容
设无穷数列{an}的各项都是正数,Sn是它的前n项之和,对于任意正整数n,an与2的等差中项等于Sn与2的等比中项,则该数列的通项公式为 (n∈N*).
分析:由等差中项和等比中项可得
=
,平方可得Sn=
,把n=1代入可得a1=2,还可得Sn-1=
,又an=SnS-n-1,数列各项都是正数,可得an-an-1=4,可得数列为等差数列,可得通项公式.
an+2 |
2 |
2Sn |
(an+2)2 |
8 |
(an-1+2)2 |
8 |
解答:解:由题意知
=
,平方可得Sn=
,①
①由a1=S1得
=
,从而可解得a1=2.
又由①式得Sn-1=
(n≥2)…②
①-②可得an=SnS-n-1=
-
(n≥2)
整理得(an+an-1)(an-an-1-4)=0
∵数列{an}的各项都是正数,
∴an-an-1-4=0,即an-an-1=4.
故数列{an}是以2为首项4为公差的等差数列,
故其通项公式为an=2+4(n-1)=4n-2,
故答案为:an=4n-2
an+2 |
2 |
2Sn |
(an+2)2 |
8 |
①由a1=S1得
a1+2 |
2 |
2a1 |
又由①式得Sn-1=
(an-1+2)2 |
8 |
①-②可得an=SnS-n-1=
(an+2)2 |
8 |
(an-1+2)2 |
8 |
整理得(an+an-1)(an-an-1-4)=0
∵数列{an}的各项都是正数,
∴an-an-1-4=0,即an-an-1=4.
故数列{an}是以2为首项4为公差的等差数列,
故其通项公式为an=2+4(n-1)=4n-2,
故答案为:an=4n-2
点评:本题考查等差数列和等比数列的性质,属中档题.

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