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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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