(Ⅱ)若.試問(wèn)數(shù)列中是否存在整數(shù).使得對(duì)任意的正整數(shù)都有成立?并證明你的結(jié)論. 河西區(qū)2008―2009學(xué)年度第二學(xué)期高三年級(jí)總復(fù)習(xí)質(zhì)量調(diào)查(二) 查看更多

 

題目列表(包括答案和解析)

已知數(shù)列中,,且點(diǎn)P在直線x-y+1=0上。
(1)求數(shù)列的通項(xiàng)公式;
(2)設(shè),求數(shù)列的前n項(xiàng)和Tn
(3)設(shè)表示數(shù)列的前n項(xiàng)和。試問(wèn):是否存在關(guān)于n的整式,使得對(duì)于一切不小于2的自然數(shù)n恒成立? 若存在,寫(xiě)出的解析式,并加以證明;若不存在,試說(shuō)明理由。

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已知數(shù)列{an}是以d為公差的等差數(shù)列,{bn}數(shù)列是以q為公比的等比數(shù)列.
(Ⅰ)若數(shù)列的前n項(xiàng)和為Sn,且a1=b1=d=2,S3<a1003+5b2-2010,求整數(shù)q的值;
(Ⅱ)在(Ⅰ)的條件下,試問(wèn)數(shù)列中是否存在一項(xiàng)bk,使得bk恰好可以表示為該數(shù)列中連續(xù)p(p∈N,p≥2)項(xiàng)的和?請(qǐng)說(shuō)明理由;
(Ⅲ)若b1=ar,b2=as≠ar,b3=at(其中t>s>r,且(s-r)是(t-r)的約數(shù)),求證:數(shù)列{bn}中每一項(xiàng)都是數(shù)列{an}中的項(xiàng).

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(1)若對(duì)于任意的n∈N*,總有
n+2
n(n+1)
=
A
n
+
B
n+1
成立,求常數(shù)A,B的值;
(2)在數(shù)列{an}中,a1=
1
2
,an=2an-1+
n+2
n(n+1)
(n≥2,n∈N*),求通項(xiàng)an;
(3)在(2)題的條件下,設(shè)bn=
n+1
2(n+1)an+2
,從數(shù)列{bn}中依次取出第k1項(xiàng),第k2項(xiàng),…第kn項(xiàng),按原來(lái)的順序組成新的數(shù)列{cn},其中cn=bkn,其中k1=m,kn+1-kn=r∈N*.試問(wèn)是否存在正整數(shù)m,r使
lim
n→+∞
(c1+c2+…+cn)=S
4
61
<S<
1
13
成立?若存在,求正整數(shù)m,r的值;不存在,說(shuō)明理由.

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已知數(shù)列{an}的前n項(xiàng)和為Sn,且滿足2Sn=pan-2n,n∈N*,其中常數(shù)p>2.
(1)證明:數(shù)列{an+1}為等比數(shù)列;
(2)若a2=3,求數(shù)列{an}的通項(xiàng)公式;
(3)對(duì)于(2)中數(shù)列{an},若數(shù)列{bn}滿足bn=log2(an+1)(n∈N*),在bk與bk+1之間插入2k-1(k∈N*)個(gè)2,得到一個(gè)新的數(shù)列{cn},試問(wèn):是否存在正整數(shù)m,使得數(shù)列{cn}的前m項(xiàng)的和Tm=2011?如果存在,求出m的值;如果不存在,說(shuō)明理由.

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已知數(shù)列{an}是以d為公差的等差數(shù)列,數(shù)列{bn}是以q為公比的等比數(shù)列.
(1)若數(shù)列{bn}的前n項(xiàng)的和為Sn,且a1=b1=d=2,S3<a1003+5b2,求整數(shù)q的值;
(2)在(1)的條件下,試問(wèn)數(shù)列{bn}中是否存在一項(xiàng)bk,使得bk恰好可以表示為該數(shù)列中連續(xù)p(p∈N,p≥2)項(xiàng)的和?請(qǐng)說(shuō)明理由;
(3)若b1=a1,b2=as≠arb3=at,(其中t>s>r,且(s-r)是(t-r)的約數(shù)),求證:數(shù)列{bn}中每一項(xiàng)都是數(shù)列{an}中的項(xiàng).

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一、選擇題:(每小題5分,共50分)

題號(hào)

1

2

3

4

5

6

7

8

9

10

答案

B

D

B

A

C

C

C

A

A

B

二、填空題:(每小題4分,共24分)

11.6ec8aac122bd4f6e     12.4       13.6ec8aac122bd4f6e      14.6ec8aac122bd4f6e     15.4   16.6ec8aac122bd4f6e

三、解答題:(共76分,以下各題為累計(jì)得分,其他解答請(qǐng)相應(yīng)給分)

17.解:(I)6ec8aac122bd4f6e

        6ec8aac122bd4f6e  6ec8aac122bd4f6e

        由6ec8aac122bd4f6e,得6ec8aac122bd4f6e。

        又當(dāng)6ec8aac122bd4f6e時(shí)6ec8aac122bd4f6e,得6ec8aac122bd4f6e

        6ec8aac122bd4f6e

       (Ⅱ)當(dāng)6ec8aac122bd4f6e

        即6ec8aac122bd4f6e時(shí)函數(shù)遞增。

        故6ec8aac122bd4f6e的單調(diào)增區(qū)間為6ec8aac122bd4f6e6ec8aac122bd4f6e

18.解:(I)各取1個(gè)球的結(jié)果有(紅,紅1)(紅,紅2)(紅,白1)(紅,白2)(紅,黑)

(白,紅2)(白,紅2)(白,白1)(白,白2)(白,黑)(白,紅1)(白,紅2

(白,白1)(白,白2)(白,黑)(黑1,紅1)(黑1,紅2)(黑1,白1)(黑1,白2)(黑1,黑)(黑2,紅1)(黑2,紅2)(黑2,白1)(黑2,白2)(黑2,黑)(黑3,紅1

(黑3,紅2)(黑3,白1)(黑3,白2)(黑3,黑)

等30種情況

其中恰有1白1黑有(白,黑)…(黑3,白2)8種情況,

故1白1黑的概率為6ec8aac122bd4f6e

   (Ⅱ)2紅有2種,2白有4種,2黑有3種,

故兩球顏色相同的概率為6ec8aac122bd4f6e

   (Ⅲ)1紅有1×3+2×5=13(種),2紅有2種,

故至少有1個(gè)紅球的概率為6ec8aac122bd4f6e

19.解:(I)側(cè)視圖6ec8aac122bd4f6e   (高4,底26ec8aac122bd4f6e

        6ec8aac122bd4f6e

   (Ⅱ)證明,由6ec8aac122bd4f6e面ABC得6ec8aac122bd4f6eAC,又由俯視圖知AB6ec8aac122bd4f6eAC,6ec8aac122bd4f6e

6ec8aac122bd4f6e面PAB

又AC6ec8aac122bd4f6e面PAC,6ec8aac122bd4f6e面PAC6ec8aac122bd4f6e面PAB

   (Ⅲ)6ec8aac122bd4f6e面ABC,6ec8aac122bd4f6e為直線PC與底面ABC所成的角

6ec8aac122bd4f6e中,PA=4,AC=6ec8aac122bd4f6e6ec8aac122bd4f6e,

6ec8aac122bd4f6e

20.解:(I)由題意設(shè)C的方程為6ec8aac122bd4f6e6ec8aac122bd4f6e,得6ec8aac122bd4f6e。

    6ec8aac122bd4f6e

    設(shè)直線6ec8aac122bd4f6e的方程為6ec8aac122bd4f6e,由6ec8aac122bd4f6e

    ②代入①化簡(jiǎn)整理得   6ec8aac122bd4f6e

    因直線6ec8aac122bd4f6e與拋物線C相交于不同的兩點(diǎn),

    故6ec8aac122bd4f6e

    即6ec8aac122bd4f6e,解得6ec8aac122bd4f6e6ec8aac122bd4f6e時(shí)僅交一點(diǎn),6ec8aac122bd4f6e

   (Ⅱ)設(shè)6ec8aac122bd4f6e,由由(I)知

    6ec8aac122bd4f6e

    6ec8aac122bd4f6e

    6ec8aac122bd4f6e

21.解:(I)6ec8aac122bd4f6e   由6ec8aac122bd4f6e6ec8aac122bd4f6e

于是6ec8aac122bd4f6e6ec8aac122bd4f6e

6ec8aac122bd4f6e切線方程為6ec8aac122bd4f6e,即6ec8aac122bd4f6e

   (Ⅱ)令6ec8aac122bd4f6e,解得6ec8aac122bd4f6e

    ①當(dāng)6ec8aac122bd4f6e時(shí),即6ec8aac122bd4f6e時(shí),在6ec8aac122bd4f6e內(nèi),6ec8aac122bd4f6e,于是6ec8aac122bd4f6e在[1,4]內(nèi)為增函數(shù)。從而6ec8aac122bd4f6e

    ②當(dāng)6ec8aac122bd4f6e,即6ec8aac122bd4f6e,在6ec8aac122bd4f6e內(nèi),6ec8aac122bd4f6e,于是6ec8aac122bd4f6e在[1,4]內(nèi)為減函數(shù),從而6ec8aac122bd4f6e

    ③當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e6ec8aac122bd4f6e內(nèi)遞減,在6ec8aac122bd4f6e內(nèi)遞增,故6ec8aac122bd4f6e在[1,4]上的最大值為6ec8aac122bd4f6e6ec8aac122bd4f6e的較大者。

    由6ec8aac122bd4f6e,得6ec8aac122bd4f6e,故當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e

    當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e

22.解:(I)設(shè)6ec8aac122bd4f6e的首項(xiàng)為6ec8aac122bd4f6e,公差為d,于是由6ec8aac122bd4f6e

        解得6ec8aac122bd4f6e        6ec8aac122bd4f6e

       (Ⅱ)6ec8aac122bd4f6e

        由6ec8aac122bd4f6e  ①

        得6ec8aac122bd4f6e     ②

        ①―②得6ec8aac122bd4f6e   即6ec8aac122bd4f6e

        當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e,當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e

        6ec8aac122bd4f6e

        于是6ec8aac122bd4f6e

        設(shè)存在正整數(shù)6ec8aac122bd4f6e,使對(duì)6ec8aac122bd4f6e恒成立

        當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e,即6ec8aac122bd4f6e

        當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e

        6ec8aac122bd4f6e

        6ec8aac122bd4f6e當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e,當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e

        6ec8aac122bd4f6e存在正整數(shù)6ec8aac122bd4f6e或8,對(duì)于任意正整數(shù)6ec8aac122bd4f6e都有6ec8aac122bd4f6e成立。

 


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