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0829xin:高中数学 第二章 数列复习课导学案新人教A版必修5

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【步步高】2014-2015 学年高中数学 第二章 数列复习课检测试题 新人教 A 版必修 5
课时目标 综合运用等差数列与等比数列的有关知识,解决数列综合问题和实际问题.

一、选择题 1.在如图的表格中,每格填上一个数字后,使每一横行成等差数列,每一纵列成等比 数列,则 a+b+c 的值为( ) 1 2 1 1 2

a b c
A.1 答案 解析 B.2 A C.3 D.4

1 5 3 由题意知,a= ,b= ,c= , 2 16 16 故 a+b+c=1. 2.已知等比数列{an},a1=3,且 4a1、2a2、a3 成等差数列,则 a3+a4+a5 等于( A.33 B.72 C.84 D.189 答案 C 解析 由题意可设公比为 q,则 4a2=4a1+a3, 又 a1=3,∴q=2. 2 2 ∴a3+a4+a5=a1q (1+q+q ) =3×4×(1+2+4)=84.

)

3.已知一个等比数列首项为 1,项数为偶数,其奇数项和为 85,偶数项之和为 170, 则这个数列的项数为( ) A.4 B.6 C.8 D.10 答案 C 解析 设项数为 2n,公比为 q. 由已知 S 奇=a1+a3+?+a2n-1. ① S 偶=a2+a4+?+a2n. ② 170 ②÷①得,q= =2, 85 a1?1-q2n? 1-22n ∴S2n=S 奇+S 偶=255= = , 1-q 1-2 ∴2n=8. 4.在公差不为零的等差数列{an}中,a1,a3,a7 依次成等比数列,前 7 项和为 35,则数 列{an}的通项 an 等于( ) A.n B.n+1 C.2n-1 D.2n+1 答案 B 2 2 解析 由题意 a3=a1a7,即(a1+2d) =a1(a1+6d), 2 得 a1d=2d . 7×6 又 d≠0,∴a1=2d,S7=7a1+ d=35d=35. 2 ∴d=1,a1=2,an=a1+(n-1)d=n+1. 5.在数列{an}中,a1=1,anan-1=an-1+(-1) (n≥2,n∈N+),则 的值是( A. 15 16 答案 解析 15 B. 8 3 C. 4 3 D. 8
n

a3 a5

)

C 2 由已知得 a2=1+(-1) =2, 1 3 ∴a3·a2=a2+(-1) ,∴a3= , 2 1 1 4 ∴ a4= +(-1) ,∴a4=3, 2 2 2 5 ∴3a5=3+(-1) ,∴a5= , 3 a3 1 3 3 ∴ = × = . a5 2 2 4 6.已知等比数列{an}的各项均为正数,数列{bn}满足 bn=ln an,b3=18,b6=12,则数 列{bn}前 n 项和的最大值等于( ) A.126 B.130 C.132 D.134 答案 C 解析 ∵{an}是各项不为 0 的正项等比数列, ∴{bn}是等差数列. 又∵b3=18,b6=12,∴b1=22,d=-2, n?n-1? 2 ∴Sn=22n+ ×(-2)=-n +23n, 2 2 23 2 23 =-(n- ) + 2 4 ∴当 n=11 或 12 时,Sn 最大, 2 ∴(Sn)max=-11 +23×11=132. 二、填空题 7.三个数成等比数列,它们的和为 14,积为 64,则这三个数按从小到大的顺序依次为

__________. 答案 2,4,8 解析 设这三个数为 ,a,aq.由 ·a·aq=a =64,得 a=4.

a q

a q

3

a 4 1 由 +a+aq= +4+4q=14.解得 q= 或 q=2. q q 2 ∴这三个数从小到大依次为 2,4,8. 8.一个等差数列的前 12 项和为 354,前 12 项中偶数项与奇数项和之比为 32∶27,则 这个等差数列的公差是____. 答案 5 解析 S 偶=a2+a4+a6+a8+a10+a12;S 奇=a1+a3+a5+a7+a9+a11. ? ?S奇+S偶=354 则? ,∴S 奇=162,S 偶=192, ?S偶÷S奇=32∶27 ? ∴S 偶-S 奇=6d=30,d=5. 9.如果 b 是 a,c 的等差中项,y 是 x 与 z 的等比中项,且 x,y,z 都是正数,则(b- c)logmx+(c-a)logmy+(a-b)logmz=______. 答案 0 解析 ∵a,b,c 成等差数列,设公差为 d, 则(b-c)logmx+(c-a)logmy+(a-b)logmz=-dlogmx+2dlogmy-dlogmz
=dlogm =dlogm1=0. 10.等比数列{an}中,S3=3,S6=9,则 a13+a14+a15=________. 答案 48

y2 xz

解析

易知 q≠1,∴

? ? ? a ?1-q ? ? ?S = 1-q =9
6 1 6

a1?1-q3? S3= =3 1-q



S6 3 3 S3 12 ∴a13+a14+a15=(a1+a2+a3)q 12 4 =S3·q =3×2 =48.
∴ =1+q =3,∴q =2. 三、解答题 21 1 ?1? 11.设{an}是等差数列,bn=? ?an,已知:b1+b2+b3= ,b1b2b3= ,求等差数列的通 2 8 8 ? ? 项 an. 解 设等差数列{an}的公差为 d, ?1?a ? ? n+1 1 bn+1 ?2? ? ? ?1?d 则 = =? ?an+1-an=? ? . bn ?2? ?1?a ?2? ?2? n ? ? ?1?d ∴数列{bn}是等比数列,公比 q=? ? . ?2? 1 1 3 ∴b1b2b3=b2= ,∴b2= . 8 2

17 b +b = ? ? 8 ∴? 1 b ·b = ? ? 4
1 3 1 3

1 ? ?b1= 8 ,解得? ? ?b3=2

b1=2 ? ? 或? 1 b3= ? 8 ?

.

1 ? ?b1= 8 当? ? ?b3=2

时,q =16,∴q=4(q=-4<0 舍去)

2

?1? n-1 2n-5 =? ?·4 =2 . ?8? ?1?5-2n ?1? 由 bn=? ? =? ?an,∴an=5-2n. ?2? ?2?
此时,bn=b1q
n-1

b1=2 ? ? 当? 1 b3= ? 8 ?

1 1 1? ? 2 时,q = ,∴q= ?q=- <0舍去? 4 16 4? ?
n-1

此时,bn=b1q

?1?n-1 ?1?2n-3 ?1? =2·? ? =? ? =? ?an, ?4? ?2? ?2?

∴an=2n-3. 综上所述,an=5-2n 或 an=2n-3. 12.已知等差数列{an}的首项 a1=1,公差 d>0,且第二项、第五项、第十四项分别是一 个等比数列的第二项、第三项、第四项. (1)求数列{an}的通项公式; 1 * (2)设 bn= (n∈N ),Sn=b1+b2+?+bn,是否存在 t,使得对任意的 n 均有 n?an+3?

t Sn> 总成立?若存在,求出最大的整数 t;若不存在,请说明理由.

36 2 2 解 (1)由题意得(a1+d)(a1+13d)=(a1+4d) ,整理得 2a1d=d .∵d>0,∴d=2 * ∵a1=1.∴an=2n-1 (n∈N ). 1 ? 1 1 1? 1 (2)bn= = = ? - ?, n?an+3? 2n?n+1? 2?n n+1? ∴Sn=b1+b2+?+bn 1 ?? 1?? 1? ?1 1? ?1 = ??1- ?+? - ?+?+? - ?? 2? ?2 3? 2?? ?n n+1?? 1 ? 1? n = ?1- = . n+1? 2? 2 ? n +1? ?

假设存在整数 t 满足 Sn> 总成立, 36 n+1 n 1 又 Sn+1-Sn= - = >0, 2?n+2? 2?n+1? 2?n+2??n+1? ∴数列{Sn}是单调递增的. 1 t 1 ∴S1= 为 Sn 的最小值,故 < ,即 t<9. 4 36 4 又∵t∈Z,∴适合条件的 t 的最大值为 8. 能力提升 13.已知数列{an}为等差数列,公差 d≠0,其中 ak1,ak2,?,akn 恰为等比数列,若 k1=1,k2=5,k3=17,求 k1+k2+?+kn. 2 解 由题意知 a5=a1a17, 2 即(a1+4d) =a1(a1+16d).

t

∵d≠0,由此解得 2d=a1. a5 a1+4d n-1 公比 q= = =3.∴akn=a1·3 .

a1

a1

又 akn=a1+(kn-1)d= ∴a1·3
n-1

kn+1 a1,
2

2 n-1 ∵a1≠0,∴kn=2·3 -1, n-1 ∴k1+k2+?+kn=2(1+3+?+3 )-n n =3 -n-1. 14.设数列{an}的首项 a1=1,前 n 项和 Sn 满足关系式: 3tSn-(2t+3)Sn-1=3t (t>0,n=2,3,4,?). (1)求证:数列{an}是等比数列; (2)设数列{an}的公比为 f(t),作数列{bn},使 b1=1,bn=f?



kn+1 a1.

? 1 ? (n=2,3,4,?).求 ? ?bn-1?

数列{bn}的通项 bn; (3)求和:b1b2-b2b3+b3b4-b4b5+?+b2n-1b2n-b2n·b2n+1. (1)证明 由 a1=S1=1,S2=1+a2, 3+2t a2 3+2t 得 a2= , = . 3t a1 3t 又 3tSn-(2t+3)Sn-1=3t, ① 3tSn-1-(2t+3)Sn-2=3t. ② ①-②,得 3tan-(2t+3)an-1=0. an 2t+3 ∴ = ,(n=2,3,?). an-1 3t ∴数列{an}是一个首项为 1, 2t+3 公比为 的等比数列. 3t 2t+3 2 1 (2)解 由 f(t)= = + , 3t 3 t 1 2 ? ?= +b . 得 bn=f? ? n-1 ?bn-1? 3 2 ∴数列{bn}是一个首项为 1,公差为 的等差数列. 3 2 2n+1 ∴bn=1+ (n-1)= . 3 3 2n+1 5 4 (3)解 由 bn= ,可知{b2n-1}和{b2n}是首项分别为 1 和 ,公差均为 的等差数列. 3 3 3 于是 b1b2-b2b3+b3b4-b4b5+?+b2n-1b2n-b2nb2n+1 =b2(b1-b3)+b4(b3-b5)+b6(b5-b7)+?+b2n(b2n-1-b2n+1) 4 4 1 ?5 4n+1? =- (b2+b4+?+b2n)=- · n? + 3 ? 3 3 2 ?3 ? 4 2 =- (2n +3n). 9

1.等差数列和等比数列各有五个量 a1,n,d,an,Sn 或 a1,n,q,an,Sn.一般可以“知 三求二”,通过列方程(组)求关键量 a1 和 d(或 q),问题可迎刃而解. 2.数列的综合问题通常可以从以下三个角度去考虑:①建立基本量的方程(组)求解;

②巧用等差数列或等比数列的性质求解;③构建递推关系求解.


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以上会员名单排名不分前后