2014届数学试题选编12：等差数列及其前n项和（教师版） Word版含(5)

123d)?a(a?d) 22112111∴ad?d?0 ∴d(a?d)?0 ∵d?0 ∴a?d ∴d?2a 24222n(n?1)n(n?1)d?na?2a?n2a ∴Sn?na?22∵b1，b2，b4成等比数列 ∴b2?b1b4 ∴(a?2∴左边=Snk?(nk)2a?n2k2a 右边=n2Sk?n2k2a ∴左边=右边∴原式成立

(2)∵{bn}是等差数列∴设公差为d1,∴bn?b1?(n?1)d1带入bn?nSn得: 2n?cb1?(n?1)d1??nSn1132(d?d)n?(b?d?a?d)n?cd1n?c(d1?b1) ∴111222n?c对n?N恒成立

1?d??12d?0??b?d?a?1d?0∴?1 12??cd1?0?c(d?b)?0?11由①式得:d1?1d ∵ d?0 ∴ d1?0 2证

:(1)

若

,

则

由③式得:c?0 法二:

c?0n[(n?1)d?2a](n?1)d?2a,bn?. an?a?(n?1)d,Sn?222当b1，b2，b4成等比数列,b2?b1b4,

d?3d???2即:?a???a?a??,得:d?2ad,又d?0,故d?2a.

2?2???由此:Sn?n2a,Snk?(nk)2a?n2k2a,n2Sk?n2k2a. 故:Snk?n2Sk(k,n?N*).

2(n?1)d?2anS2(2)bn?2n?, 2n?cn?c(n?1)d?2a(n?1)d?2a(n?1)d?2an2?c?c222 ?2n?cn2

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(n?1)d?2a(n?1)d?2a2. (※) ??22n?cc若{bn}是等差数列,则bn?An?Bn型. 观察(※)式后一项,分子幂低于分母幂,

(n?1)d?2a(n?1)d?2a(n?1)d?2a2c?0故有:,即,而≠0, ?0222n?c故c?0.

c经检验,当c?0时{bn}是等差数列.

24．（江苏省南京市四校2013届高三上学期期中联考数学试题）已知等差数列

?an?的前n项和

为Sn,公差d?0,且S3?S5?50,a1,a4,a13成等比数列. (Ⅰ)求数列?an?的通项公式; (Ⅱ)设?

【答案】解:(Ⅰ)依题意得

?bn??是首项为1,公比为3的等比数列,求数列?bn?的前n项和Tn. a?n?3?24?5?d?5a1?d?50?3a1? 22??(a?3d)2?a(a?12d)11?1解得??a1?3,

?d?2?an?a1?(n?1)d?3?2(n?1)?2n?1，即an?2n?1

(Ⅱ)

bn?3n?1,bn?an?3n?1?(2n?1)?3n?1 anTn?3?5?3?7?32???(2n?1)?3n?1

3Tn?3?3?5?32?7?33???(2n?1)?3n?1?(2n?1)?3n

?2Tn?3?2?3?2?32???2?3n?1?(2n?1)3n

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3(1?3n?1)?3?2??(2n?1)3n 1?3??2n?3n∴Tn?n?3n

25．（扬州市2012-2013学年度第一学期期末检测高三数学试题）已知数列{an}的前n项和为

Sn.

(Ⅰ)若数列{an}是等比数列,满足2a1列

?a3?3a2, a3?2是a2,a4的等差中项,求数

?an?的通项公式;

(Ⅱ)是否存在等差数列{an},使对任意n?N*都有an?Sn?2n2(n?1)?若存在,请求出所有满足条件的等差数列;若不存在,请说明理由.

【答案】解:(Ⅰ)设等比数列

?an?的首项为a1,公比为q,

?a1(2?q2)?3a1q,(1)?2a1?a3?3a2,依题意,有?即?32a?a?2(a?2).43?2?a1(q?q)?2a1q?4.(2)由 (1)得 q?3q?2?0,解得q?1或q2?2.

当q当q?1时,不合题意舍;

?2时,代入(2)得a1?2,所以,an?2?2n?1?2n

(Ⅱ)假设存在满足条件的数列{an},设此数列的公差为d,则 方法1: [a1?(n?1)d][a1n?n(n?1)d]?2n2(n?1),得 2d22331n?(a1d?d2)n?(a12?a1d?d2)?2n2?2n对n?N*恒成立, 2222?d2?2?2,??32则?a1d?d?2,

2?12?23a?ad?d?0,?1212?解得?

?d?2,?d??2,或?此时an?2n,或an??2n.

?a1?2,?a1??2.18

故存在等差数列{an},使对任意n?N*都有an?Sn?2n2(n?1).其中an?2n, 或an??2n

方法2:令n?1,a12?4,得a1??2,

2令n?2,得a2?a1?a2?24?0,

①当a1?2时,得a2?4或a2??6,

若a2?4,则d?2,an?2n,Sn?n(n?1),对任意n?N*都有

an?Sn?2n2(n?1);

若a2??6,则d??8,a3??14,S3??18,不满足a3?S3?2?32?(3?1). ②当a1??2时,得a2??4或a2?6,

若a2??4,则d??2,an??2n,Sn??n(n?1),对任意n?N*都有

an?Sn?2n2(n?1);

若a2?6,则d?8,a3?14,S3?18,不满足a3?S3?2?32?(3?1).

综上所述,存在等差数列{an},使对任意n?N*都有an?Sn?2n2(n?1).其中

an?2n,或an??2n

26．（苏州市2012-2013学年度第一学期高三期末考试数学试卷）

设数列?an?的前n项和为Sn,满足an?Sn?An2?Bn?1(A?0).

39,a2?,求证数列?an?n?是等比数列,并求数列?an?的通项公式; 24B?1(2)已知数列?an?是等差数列,求的值.

A(1)若a1?

【答案】

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27．（2012年江苏理）已知各项均为正数的两个数列{an}和{bn}满

足:an?1?an?bnan?bn22,n?N*,

(1)设bn?1?bn??b??1?,n?N*,求证:数列??n?aan???n?2???是等差数列; ??(2)设bn?1?2?bn,n?N*,且{an}是等比数列,求a1和b1的值. an【答案】解:(1)∵bn?1?1?bnan?bn,∴an?1?=22anan?bnbn?1?b?1??n??an?2. 20

∴

bn?1an?12?2??bn??bn?1??bn???bn???bn??1???.∴ ???????1????????1?n?N*? .

aa?n??n?1??an???an???an???22222???bn???∴数列????是以1 为公差的等差数列.

a???n???(2)∵an>0，bn>0?a?bn?,∴n22?an2?bn2

2∴1

设等比数列{an}的公比为q,由an>0知q>0,下面用反证法证明q=1 若q>1,则a1=a22logq时,an?1?a1qn>2,与(﹡)矛盾. qa1a21>a2>1,∴当n>logq时,an?1?a1qn<1,与(﹡)矛盾. qa1若0

∴b1，b2，b3中至少有两项相同,与b1

∴bn=2??2??2?22?2?2?2=2. ?1∴ a1=b2=2.

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