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-.z.高中數(shù)學(xué)數(shù)列壓軸題練習(xí)〔〕及詳解1.數(shù)列是公差為正數(shù)的等差數(shù)列,其前n項和為,且?,

(Ⅰ)求數(shù)列的通項公式;

(Ⅱ)數(shù)列滿足,①求數(shù)列的通項公式;

②是否存在正整數(shù)m,,使得,,成等差數(shù)列"假設(shè)存在,求出m,n的值;假設(shè)不存在,請說明理由.解:(I)設(shè)數(shù)列的公差為d,則由?,,得,

計算得出或(舍去).

;

(Ⅱ)①,,

,

,

即,,,,累加得:,

也符合上式.

故,.

②假設(shè)存在正整數(shù)m、,使得,,成等差數(shù)列,

則又,,,

,即,

化簡得:當(dāng),即時,,(舍去);

當(dāng),即時,,符合題意.

存在正整數(shù),,使得,,成等差數(shù)列.解析(Ⅰ)直接由列關(guān)于首項和公差的方程組,求解方程組得首項和公差,代入等差數(shù)列的通項公式得答案;

(Ⅱ)①把數(shù)列的通項公式代入,然后裂項,累加后即可求得數(shù)列的通項公式;

②假設(shè)存在正整數(shù)m、,使得,,成等差數(shù)列,則.由此列關(guān)于m的方程,求計算得出答案.2.在數(shù)列中,,

(1)求證:數(shù)列為等比數(shù)列;

(2)記,且數(shù)列的前n項和為,假設(shè)為數(shù)列中的最小項,求的取值范圍.解:(1)證明:,

又,

,,

故,

是以3為首項,公比為3的等比數(shù)列

(2)由(1)知道,,

假設(shè)為數(shù)列中的最小項,則對有恒成立,

即對恒成立當(dāng)時,有;

當(dāng)時,有?;

當(dāng)時,恒成立,

對恒成立.

令,則對恒成立,

在時為單調(diào)遞增數(shù)列.

,即綜上,解析(1)由,整理得:.由,,可以知道是以3為首項,公比為3的等比數(shù)列;

(2)由(1)求得數(shù)列通項公式及前n項和為,由為數(shù)列中的最小項,則對有恒成立,分類分別求得當(dāng)時和當(dāng)?shù)娜≈捣秶?

當(dāng)時,,利用做差法,根據(jù)函數(shù)的單調(diào)性,即可求得的取值范圍.3.在數(shù)列中,

,

,

,設(shè)為的前n項和.

(1)求證:數(shù)列是等差數(shù)列;

(2)求;

(3)是否存在正整數(shù)p,q,

,使,

,

成等差數(shù)列"假設(shè)存在,求出p,q,r的值;假設(shè)不存在,說明理由.(1)證明:由,,

得到,

則又,

,

數(shù)列是以1為首項,以-2為公差的等差數(shù)列;

(2)由(1)可以推知:,

所以,,

所以,①,②①-②,得,

,

,

所以

(3)假設(shè)存在正整數(shù)p,q,,使,,成等差數(shù)列.

則,

即因為當(dāng)時,,

所以數(shù)列單調(diào)遞減.

又,

所以且q至少為2,

所以,①當(dāng)時,,

又,

所以,等式不成立.

②當(dāng)時,,

所以所以,

所以,(數(shù)列單調(diào)遞減,解唯一確定).

綜上可以知道,p,q,r的值分別是1,2,3.解析(1)把給出的數(shù)列遞推式,,變形后得到新數(shù)列,該數(shù)列是以1為首項,以-2為公差的等差數(shù)列;

(2)由(1)推出的通項公式,利用錯位相減法從而求得求;

(3)根據(jù)等差數(shù)列的性質(zhì)得到,從而推知p,q,r的值.4.n為正整數(shù),數(shù)列滿足,

,設(shè)數(shù)列滿足

(1)求證:數(shù)列為等比數(shù)列;

(2)假設(shè)數(shù)列是等差數(shù)列,求實數(shù)t的值;

(3)假設(shè)數(shù)列是等差數(shù)列,前n項和為,對任意的,均存在,使得成立,求滿足條件的所有整數(shù)的值.(1)證明:數(shù)列滿足,,

?,?,

數(shù)列為等比數(shù)列,其首項為,公比為2;

(2)解:由(1)可得:?,

,數(shù)列是等差數(shù)列,,

,

計算得出或12.

時,,是關(guān)于n的一次函數(shù),因此數(shù)列是等差數(shù)列.

時,,,不是關(guān)于n的一次函數(shù),

因此數(shù)列不是等差數(shù)列.

綜上可得;

(3)解:由(2)得,

對任意的,均存在,使得成立,

即有??,

化簡可得,

當(dāng),,,對任意的,符合題意;

當(dāng),,當(dāng)時,,

對任意的,不符合題意.

綜上可得,當(dāng),,對任意的,均存在,

使得成立.解析(1)根據(jù)題意整理可得,?,再由等比數(shù)列的定義即可得證;

(2)運用等比數(shù)列的通項公式和等差數(shù)列中項的性質(zhì),可得,解方程可得t,對t的值,檢驗即可得到所求值;

(3)由(2)可得,對任意的,均存在,使得成立,即有??,討論為偶數(shù)和奇數(shù),化簡整理,即可得到所求值.5.常數(shù),數(shù)列滿足,

(1)假設(shè),

,

①求的值;

②求數(shù)列的前n項和;

(2)假設(shè)數(shù)列中存在三項,

,

依次成等差數(shù)列,求的取值范圍.解:(1)①,

,

,

,

②,,

當(dāng)時,,

當(dāng)時,,即從第二項起,數(shù)列是以1為首項,以3為公比的等比數(shù)列,

數(shù)列的前n項和,,

顯然當(dāng)時,上式也成立,

;

(2),

,即單調(diào)遞增.

(i)當(dāng)時,有,于是,

,假設(shè)數(shù)列中存在三項,,依次成等差數(shù)列,則有,

即,.因此不成立.因此此時數(shù)列中不存在三項,,依次成等差數(shù)列.

當(dāng)時,有.此時于是當(dāng)時,.從而假設(shè)數(shù)列中存在三項,,依次成等差數(shù)列,則有,

同(i)可以知道:.于是有,,是整數(shù),.于是,即.與矛盾.

故此時數(shù)列中不存在三項,,依次成等差數(shù)列.

當(dāng)時,有于是此時數(shù)列中存在三項,,依次成等差數(shù)列.

綜上可得:解析(1)①,可得,同理可得,②,,當(dāng)時,,當(dāng)時,,即從第二項起,數(shù)列是以1為首項,以3為公比的等比數(shù)列,利用等比數(shù)列的求和公式即可得出

(2),可得,即單調(diào)遞增.

(i)當(dāng)時,有,于是,可得,.利用反證法即可得出不存在.

當(dāng)時,有.此時.于是當(dāng)時,.從而.假設(shè)存在,同(i)可以知道:.得出矛盾,因此不存在.

當(dāng)時,有.于是.即可得出結(jié)論.6.兩個無窮數(shù)列和的前n項和分別為,

,

,,對任意的,都有

(1)求數(shù)列的通項公式;

(2)假設(shè)為等差數(shù)列,對任意的,都有.證明:

;

(3)假設(shè)為等比數(shù)列,

,

,求滿足的n值.解:(1)由,得,

即,所以由,,可以知道所以數(shù)列是以1為首項,2為公差的等差數(shù)列.

故的通項公式為,

(2)證法一:設(shè)數(shù)列的公差為d,

則,

由(1)知,因為,所以,

即恒成立,

所以,即,

又由,得,

所以所以,得證.

證法二:設(shè)的公差為d,假設(shè)存在自然數(shù),使得,

則,即,

因為,所以所以,

因為,所以存在,當(dāng)時,恒成立.

這與"對任意的,都有〞矛盾!

所以,得證.

(3)由(1)知,.因為為等比數(shù)列,

且,,

所以是以1為首項,3為公比的等比數(shù)列.

所以,則,

因為,所以,所以而,所以,即當(dāng),2時,式成立;

當(dāng)時,設(shè),

則,

所以,

故滿足條件的n的值為1和2.解析(1)運用數(shù)列的遞推式和等差數(shù)列的定義和通項公式,即可得到所求;

(2)方法一、設(shè)數(shù)列的公差為d,求出,.由恒成立思想可得,求出,判斷符號即可得證;

方法二、運用反證法證明,設(shè)的公差為d,假設(shè)存在自然數(shù),使得,推理可得,作差,推出大于0,即可得證;

(3)運用等差數(shù)列和等比數(shù)列的求和公式,求得,,化簡,推出小于3,結(jié)合等差數(shù)列的通項公式和數(shù)列的單調(diào)性,即可得到所求值.7.數(shù)列,

都是單調(diào)遞增數(shù)列,假設(shè)將這兩個數(shù)列的項按由小到大的順序排成一列(一樣的項視為一項),則得到一個新數(shù)列

(1)設(shè)數(shù)列,

分別為等差、等比數(shù)列,假設(shè),

,

,求;

(2)設(shè)的首項為1,各項為正整數(shù),

,假設(shè)新數(shù)列是等差數(shù)列,求數(shù)列的前n項和;

(3)設(shè)是不小于2的正整數(shù)),

,是否存在等差數(shù)列,使得對任意的,在與之間數(shù)列的項數(shù)總是假設(shè)存在,請給出一個滿足題意的等差數(shù)列;假設(shè)不存在,請說明理由.解:(1)設(shè)等差數(shù)列的公差為d,等比數(shù)列的公比為q,

根據(jù)題意得,,計算得出或3,因數(shù)列,單調(diào)遞增,

所以,,

所以,,

所以,因為,,,

(2)設(shè)等差數(shù)列的公差為d,又,且,

所以,所以因為是中的項,所以設(shè),即當(dāng)時,計算得出,不滿足各項為正整數(shù);當(dāng)時,,此時,只需取,而等比數(shù)列的項都是等差數(shù)列,中的項,所以;當(dāng)時,,此時,只需取,

由,得,是奇數(shù),是正偶數(shù),m有正整數(shù)解,

所以等比數(shù)列的項都是等差數(shù)列中的項,所以綜上所述,數(shù)列的前n項和,或

(3)存在等差數(shù)列,只需首項,公差下證與之間數(shù)列的項數(shù)為.即證對任意正整數(shù)n,都有,

即成立.

由,

所以首項,公差的等差數(shù)列符合題意解析(1)設(shè)等差數(shù)列的公差為d,等比數(shù)列的公比為q,根據(jù)題意得,,計算得出或3,因數(shù)列,單調(diào)遞增,,,可得,,利用通項公式即可得出.

(2)設(shè)等差數(shù)列的公差為d,又,且,所以,所以.因為是中的項,所以設(shè),即.當(dāng)時,計算得出,不滿足各項為正整數(shù)當(dāng)時,當(dāng)時,即可得出.

(3)存在等差數(shù)列,只需首項,公差.下證與之間數(shù)列的項數(shù)為.即證對任意正整數(shù)n,都有,作差利用通項公式即可得出.8.對于數(shù)列,稱(其中,為數(shù)列的前k項"波動均值〞.假設(shè)對任意的,,都有,則稱數(shù)列為"趨穩(wěn)數(shù)列〞.

(1)假設(shè)數(shù)列1,*,2為"趨穩(wěn)數(shù)列〞,求*的取值范圍;

(2)假設(shè)各項均為正數(shù)的等比數(shù)列的公比,求證:是"趨穩(wěn)數(shù)列〞;

(3)數(shù)列的首項為1,各項均為整數(shù),前k項的和為.且對任意,,都有,試計算:.解:(1)根據(jù)題意可得,

即,兩邊平方可得,

計算得出;

(2)證明:由,設(shè),

因且,

故對任意的,,都有,

,,

因,

,,,,,

,

,

,

即對任意的,,都有,故是"趨穩(wěn)數(shù)列〞;

(3)當(dāng)時,當(dāng)時,,

同理,,

因,

,

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