高中数学数列专
题
快递公司问题件快递公司问题件货款处理关于圆的周长面积重点题型关于解方程组的题及答案关于南海问题
一、选择题(本大题共12小题,每小题5分,共60分,在每小题给出的四个选项中,只有
一项是符合题目要求的.)
a1( 某数列的前四项为,则以下各式 0,2,0,2,,n
,(n为偶数)2n2,n,,? ? ? a,,,11a,,,1(1)a,,,,nnn,,(n为奇数)20,,
a其中可作为的通项公式的是()D( ,,n
A(??? B(?? C(?? D(?
2(若、、成等差数列,则( ) lgalgblgc
ac,1A( B( b,bab,,lglg,,22
acacC(、 、 成等差数列 D(、 、 成等比数列 bb
23(若af(x),ax,bx,c、b、c成等差数列,则函数的图像与x轴的交点的个数是( )
A(0个 B(1个 C(2个 D(不确定
,,a,aa,a,a,?,aa4(差数列d中,公差,1,,8,则, ( ) n41724620
A(40 B(45 C(50 D(55
an,,2495(已知数列{a }的通项公式是,则S 达到最小值时,n的值是 ( ) n nn
A(23 B(24 C(25 D(26
{a}中a,0,a,0,且a,|a|6(在等差数列,则在S中最大的负数为 ( ) nn10111110
A(S B(S C(S D(S17181920
aaaa,,,39aaa,,,277((2004湖北八校联考)等差数列中,,,则数列,,n147369aS前9项和等于( ) ,,n9
A(66 B(99 C(144 D(297
1aa,8(数列的通项公式是,若前n项和为10,则项数n为( ) ,,nnnn,,1
A(11 B(99 C(120 D(121
,,,,a,0,且aaa9(数列中,是公比为的等比数列,满足 ( ) q(q,0)nnnn,1
,aa(n,N)aa,aa,则公比q的取值范围是 ( ) nn,1n,1n,2n,2n,3
1,51,20,q,0,q, A( B( 22
,1,2,1,5 C( D( 0,q,0,q,22
Sn,,1446SaS,36S,32410(设是数列的前n项和,已知,,,则n,,,,,,n,6nn6n
等于( )
A(15 B(16 C(17 D(18
11(数列{a}中,已知S =1, S=2 ,且S,3S +2S =0(n?N*),则此数列为 ( ) ,n12n+1nn1
A(等差数列 B(等比数列
C(从第二项起为等差数列 D(从第二项起为等比数列
n,79a12、已知,(),则在数列,,的前50项中最小项和最大项分别是( ) n,Na,n,nn,80
a,aa,aa,aa,a A( B( C( D( 8995015018
二、填空题:
2,,a13(数列的前,项的和S=3n+ n+1,则此数列的通项公式a=_______. n nn
114(在之间插入n个正数,使这n+2个正数成等比数列,则插入的n个正数之积和n,1n
为 .
a,a,a139,,a15(等差数列中,公差d?0,a,a,a 成等比数列,则= ____ . 13 9na,a,a2410
2n,116(当x?1,0时,1+3x+5x +„„+(2n,1)x = ___________________. 三、解答题:
aaanannn,,,,,,,2312?a17(数列满足,求 ,,,,123nn
2xnxxx,1,x18(数列中,,,求数列的通项公式 ,,,,n1n,1n2,x2n
19(某产品按质量分10个档次,生产最低档次的利润是8元/件;每提高一个档次,利润每件增加2元,每提高一个档次,产量减少3件,在相同时间内,最低档次的产品可生产60件(问:在相同时间内,生产第几档次的产品可获得最大利润,(最低档次为第一档次)
52a20(设有数列aaa,,若以,,„,为系数的二次方程:axax,,,10,,a,n12nnn,116
*,(且)都有根、满足 n,2nN,,331,,,,,,,
1,,(1)求证为等比数列; a,,,n2,,
a(2)求; n
aS(3)求的前n项和( nn
*ba,0aa21(项数都是41n, 的等差数列与等比数列的首项均为 ,,,,,,,nN,,,nn
且它们的末项相等,试比较中间项的大小(
22((本题满分14分)
x为f(x)x,R,使f(x),x 对于函数,若存在成立,则称的不动点.如果函数 f(x)0000
2x,a1f(x),(b,c,N)有且只有两个不动点0,2,且 f(,2),,,bx,c2
(1)求函数的解析式; f(x)
1aa满足Sf{}4,(),1 (2)已知各项不为零的数列,求数列通项; nnnan
a,3{a}a,4,a,f(a)n,2 (3)如果数列满足,求证:当时,恒有成立. n1n,1nn
参考
答案
八年级地理上册填图题岩土工程勘察试题省略号的作用及举例应急救援安全知识车间5s试题及答案
一、选择题
题1 2 3 4 5 6 7 8 9 10 11 12 号
答A D D B B C B C A D D C 案
二、填空题
n,n,15(n,1)2,13、 14、 ()a,,nn,6n,2(n,2),
1nn,1,x,(2n,1)x,(2n,1)x1315、 16、 2(1,x)16
三
a,,,,123617([解析]: 1
aaanannn,,,,,,,2312? 当n,2时 ? ? ,,,,123n
aaanannn,,,,,,,,23111? ? ,,,,,,1231n,
nann,,31an,,31 ?,?得 ? ,,,,nn
a,,,236n,1 当时 上式 1
an,,31 ? . ,,n
xxxx18([解析]思路1:计算出,,,猜想,再证明( 234n
22x2x2nn,x 思路2:? ? , x,1n,n122,x2,x2nn
2x,2111111n,, ? 即 ,,,22222xx2xxx22nn,1,nnn1
,,111,1 ? 数列是首项为,公差为的等差数列 ,,22xx21n,,
11111n,,,,,,,,,nn111 ? ,,,,22xx222n1
2x,0x, 由已知可得 ? nnn,119([解析]10个档次的产品的每件利润构成等差数列:8,10,12,„,
ann,,,,,82126 ,,n
110,,n,10个档次的产品相同时间内的产量构成数列:60,57,54,„,,,
bnn,,,,,6031633110,,n ,,,,n
? 在相同时间内,生产第n个档次的产品获得的利润
ynn,,,26633 ,,,,
2 ( ,,,,,696144n,,
y,,,6144864 当时 (元) n,9max
? 生产低9档次的产品可获得最大利润(
a1n20( [解析](1)证明:? , 代入 ,,,,,,,331,,,,,,,aan,1n,1
11 得 aa,,nn,133
1111a,,a,nn13322,, ? 为定值 113aa,,nn,,1122
1,, ? 数列是等比数列 a,,,n2,,
1511(2)? a,,,,12623
nn,11111,,,, ? a,,,,,,,,n2333,,,,
n11,, ? a,,,,n32,,
111n,,(3) S,,,,,,,,n,,2n3332,,
11,,1,,,nn33,,,, 12,13
n,11 ,,n223,
ba21([解析]设的公差为,的公比为,则它们的中间项分别为 dq,,,,nn
21n,aand,,,(21) , baq,2n2n
ab, 由得 4141nn,,
2(21)n,andaq,,,221 ,,
221n,aq,,2221,,,,,,anda ? ,,,,a
2b1222n2aa,, ? 即 aba,,,,2nnn22aa2
122 ? abbab,,,,,,nnnn2222a2
12 ,,,ba0,,n2a2
ba,当且仅当,即时,上式等号成立( q,1n
ab,ab,故当时,,当时,( q,1q,122nn22nn
ab[评析]将用
表
关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf
达是解答本题的关键;作差后的配方是判断符号的需要,也体现了“集2n2n
中变量”这一重要的数学思想(
22、(本小题满分14分)
c,2,0,,,2,x,a,1,b2,x(1,b)x,cx,a,0,解:设得:由违达定理得: ,bx,ca,2,0,,,1,b,
a,0,2x,21,f(x),解得代入表达式,由 f(,2),,,,,,cc1,c2b,1,,(1,)x,c2,2得不止有两个不动点, c,3,又c,N,b,N,若c,0,b,1,则f(x),x
2x?c,2,b,2,于是f(x),,(x,1).„„„„„„„„„„„„„„„5分 2(x,1)
12()a2n4S,,1得:2S,a,a,(2)由题设得 (A) nnnn12(,1)an
2且 (B) a,1,以n,1代n得:2S,a,ann,n,n,11122由(A)(B)得: 2a,(a,a),(a,a)即(a,a)(a,a,1),0,,nnn,1nn,1nn,1nn,1
2 ?a,,a或a,a,,1,以n,1代入(A)得:2a,a,a,nn,nn,11111
a,,a得a,1,a,1解得(舍去)或;由,若这与矛a,0a,,1a,,1nn,12n111
盾,
?a,a,,1a},即{是以1为首项,1为公差的等差数列, ,,nn,1n
?a,,n; „„„„„„„„„„„„„„„„„„„„„„„„10分 n
2ana,3(n,2),(3)证法(一):运用反证法,假设则由(1)知 a,f(a),nn,n12a,2naa11113n,1n ?,,,(1,),(1,),,1,即a,a(n,2,n,N)n,1na2(a,1)2a,1224nnn2a1681a,a,?,a?,而当 n,2时,a,,,,3;?a,3,nn,122n2a,28,231
a,3这与假设矛盾,故假设不成立,?.„„„„„„„„„„„„„„„14分 n
2a11111n2af(a)得a,2()证法(二):由 ,,,,,,,n,nn,112a2aa222,nn,n1
a,2,若a,0,则a,0,3,a得<0或结论成立; n,1n,1n,1n,1
aa,(,2)nnaaa,,,0,若,此时从而 ,2n,2,n,1n,1na2(,1)n
22an,2即数列{}在时单调递减,由,可知上成2a,a,a,2,3,在n,2n2n233立.„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„14分