?a(1?an)n(n?1)?(a?1)??1?a2 ??2?n?n(a?1)??22【第4练】
1.C x2?(2?1)(2?1)?1,x??1
2.B x(3x?3)?(2x?2)2,x??1或x??4,而x??1?x??4 q?3x?3313?,?13??4?()n?1,n?4 2x?2222321?q3313.C a1(1?q)?18,a1(q?q)?12,?,q?或q?2,
q?q2222(1?28)?29?2?510 而q?Z,q?2,a1?2,S8?1?2965a52a5a1?a92(a1?a9)S97?9?2654. ??????12b52b5b1?b99(b?b)S\99?3121925. ?7533 q6?25,q??35,a10?a9?q??7535
6．解：设原三数为3t,4t,5t,(t?0)，不妨设t?0,则(3t?1)5t?16t2,t?5
? 3t?15,t420t,?5∴原三数为215,20,。2
1n(n?1) 27．解：记Sn?1?2x?3x2?...?nxn?1,当x?1时，Sn?1?2?3?...?n?当x?1时，xSn?x?2x2?3x3?...?(n?1)xn?1?nxn,
(1?x)Sn?1?x?x?x?...?x?1?xnn?nx(x?1)??1?x∴原式=?
?n(n?1)(x?1)??223n?11?xn?nxn ?nx,Sn?1?xn?11?2n,n?5n8． 解：bn?an??，当n?5时，Sn?(9?11?2n)?10n?n2
2?2n?11,n?6
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当n?6时，Sn?S5?Sn?5?25?2???n?10n,(n?5)∴Sn??2
??n?10n?50,(n?6)n?5(1?2n?11)?n2?10n?50 2【第5练】
1.B a1a4?a32,(a2?2)(a2?4)?(a2?2)2,2a2??12,a2??6 2.A
S99a595????1 S55a3593.D lg2?lg(2x?3)?2lg(2x?1),2(2x?3)?(2x?1)2
x2 (2)?4?x2?5?0x,2?x5?,2 log54. 38 a3?a5?a2?a?638 5.an?
6． 解：Sn?3?2n,Sn?1?3?2n?1,an?Sn?Sn?1?2n?1(n?2)
72(10n?1) 9,99,999,9919?9...1?0913,?10471?,10?1,? 1091,79?5,(n?1)而a1?S1?5，∴an??n?1
?2,(n?2)7．解：设此数列的公比为q,(q?1)，项数为2n，
a2(1?q2n)1?(q2)n则S奇??85,S偶??170,
1?q21?q2a21?22n??q?2,?85,22n?256,2n?8, S奇a11?4∴q?2,项数为8
8．解：a1a3?a22?36,a2(1?q2)?60,a2?0,a2?6,1?q2?10,q??3,
2(1?3n)?400,3n?401,n?6,n?N； 当q?3时，a1?2,Sn?1?3S偶?2[1?(?3)n]当q??3时，a1??2,Sn??400,(?3)n?801,n?8,n为偶数；
1?(?3) - 22 -
∴n?8,且n为偶数
【第6练】 1.B an?1n?n?1?n?1?n,Sn?2?1?3?2?...?n?1?n
Sn?n?1?1?9,n?1?10,n?99
2.A S4?1,S8?S4?3,而S4,S8?S4,S12?S8,S16?S12,S20?S16,成等差数列 即1,3,5,7,9,a17?a18?a19?a20?S20?S16?9
3.D a5?2a4?a3?2a2?0,a5?a3?2a4?2a2,a3(q2?1)?2a2(q2?1) a3?2a2或q2?1?0,q?2,1或?1，当q?1时，an?6；
当q??1时，a1??6,an??6?(?1)n?1?6?(?1)n?2；
当q?2时，a1?3,an?3?2n?1?6?2n?2
2(m?n,0，即)4．0 Sn?anSm?n?0 ?b该二次函数经过n172,a191?a7?79,d?7,ak?a,?9k?732 13?7?(k?9)?,k?18
35．18 3a7?17a,7?d( 9)4n?11?4nnn?1n?12n?126． Sn?2?1,Sn?1?2?1,an?2,an?4,a1?1,q?4,Sn?
1?43
7． 解：显然q?1，若q?1则S3?S6?9a1,而2S9?18a1,与S3?S6?2S9矛盾
a1(1?q3)a1(1?q6)2a1(1?q9)由S3?S6?2S9? ??1?q1?q1?q12q9?q6?q3?0,2(q3)2?q3?1?0,得q3??,或q3?1,
23而q?1，∴q??4 2 - 23 -
?n?(?4),n为偶数???2n,n为偶数?2，Sn??，8． 解：Sn??
?2n?1,n为奇数?n?1?(?4)?4n?3,n为奇数??2 S15?29,S22??44,S31?61,
S15?S22?S31??76
【第7练】
1．C am?am?am2?0,am(am?2)?0,am?2,
2m?1(a1?a2m?1)?(2m?1)a2m?38,2m?1?19 22n?1(a1?a2n?1)San2an2(2n?1)2n?12．B ? ?2?2n?1??2n?1bn2bn(b1?b2n?1)T2n?13(2n?1)?13n?12 S2m?1??1?11111113．? ??1,???1,?1,??是以为首项，以?1为
na1anan?1an?1ana1?an?公差的等差数列，
11??1?(n?1)?(?1)??n,an?? ann4．100 a8?a9?a10?a11?a12?S12?S7?122?12?1?(72?7?1)?100
25．4:1:(?2) a?c?2b,c?2b?a,a?b22c?(2?b)a,a?5a?2b4 ?b0 a?b,a?4b,c??2b
100(a?)?045,a?a1?0.9,?1a10100a?a1a?9a925050 S\?(a1?a99)??0.4?10
226． 10 S100??1d?0.4,100
7．156 a3?a7?a10?a11?a4?12,a3?a11?a10?a4,a7?12,S13?8．13(a1?a13)?13a7 2?1?55?1 设an?an?1?an?2?qan?q2an,q2?q?1?0,q?0,q? 224??1?2d?q?21，9．解：（Ⅰ）设?an?的公差为d，则依题意有q?0且? ?bn?的公比为q，21?4d?q?13，??解得d?2，q?2．
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所以an?1?(n?1)d?2n?1，
bn?qn?1?2n?1． （Ⅱ）
352n?32n?1an2n?1?n?1．Sn?1?1?2???n?2?n?1，①
2222bn252n?32n?1???n?3?n?2，② 2222222n?1②－①得Sn?2?2??2???n?2?n?1，
22222Sn?2?3?1n?12n?32n?11?2n?1?112?n?1?6?n?1． ?2?2??1??2???n?2??n?1?2?2?122?22?221?2【第8练】
1?1、解析：由题设得0＜2α＜π，0≤
∴－
?3≤
π. 6??ππ≤－≤0.∴－＜2α－＜π. 6363答案：D
2、解析：可设a=1，b=2，
1?45a2?b2a?b32ab4则=，ab=2， =，===2.5.
22222a?b3a2?b22aba?b答案：≤ab≤≤
2a?b23、解析：p=a－2+
1+2≥4，而－a2+4a－2=－（a－2）2+2＜2，∴q＜4.∴p＞q. a?2答案：A
4、解析：令f（x）=ax2+bx+c，其图象如下图所示，
y y = f( )x--3 -2 Oy = fx()2 3x 再画出f（－x）的图象即可.
答案：{x|－3＜x＜－2}
5、解析：由题意，知0、2是方程－x2+（2－m）x=0的两个根，
∴－
2?m=0+2.∴m=1. 1?212 - 25 -