郑学伟 第 13 页 2013-4-15
10.(湖南卷18).(本小题满分12分) 数列?an?满足a1?1,a2?2,an?2?(1?cos2 (Ⅰ)求a3,a4,并求数列?an?的通项公式; (Ⅱ)设bn?a2n?1a2nSn?2?,Sn?b1?b2???bn.证明:当n?6时,1n.
n?2)an?sin2n?2,n?1,2,3,?.
解: (Ⅰ)因为a1?1,a2?2,所以a3?(1?cos2?2)a1?sin2?2?a1?1?2,
a4?(1?cos2?)a2?sin2??2a2?4.
一般地,当n?2k?1(k?N*)时,a2k?1?[1?cos=a2k?1?1,即a2k?1?a2k?1?1.
所以数列?a2k?1?是首项为1、公差为1的等差数列,因此a2k?1?k. 当n?2k(k?N*)时,a2k?2?(1?cos22(2k?1)?2]a2k?1?sin22k?12?
2k?2)a2k?sin22k?2?2a2k.
k所以数列?a2k?是首项为2、公比为2的等比数列,因此a2k?2.
?n?1*,n?2k?1(k?N),?2故数列?an?的通项公式为an??
n?2*?2,n?2k(k?N).(Ⅱ)由(Ⅰ)知,bn?a2n?1a2n?n22,Sn?1212122?222?323???n2n, ① nSn??122?2?n324???2n?1 ②
①-②得,
12Sn?12?122?123???2n2n?1.
12[1?()]n1n2 ?2?n?1?1?n?n?1.
12221?21 所以Sn?2?12n?1?n2n?2?n?221nn.
n(n?2)2n 要证明当n?6时,Sn?2? 证法一
成立,只需证明当n?6时,?1成立.
延津县高级中学 z.hxw.2008@163.com
郑学伟 第 14 页 2013-4-15
(1)当n = 6时,
6?(6?2)26?4864?34?1成立.
k(k?2)2k (2)假设当n?k(k?6)时不等式成立,即
(k?1)(k?3)2k?1?1.
则当n=k+1时,
?k(k?2)2k?(k?1)(k?3)2k(k?2)?(k?1)(k?3)(k?2)?2k1n.
?1.
由(1)、(2)所述,当n≥6时, 证法二 令cn?n(n?2)22n(n?1)22?1.即当n≥6时,Sn?2?(n?6),则cn?1?cn?(n?1)(n?3)2n?1?n(n?2)26?8642?3?n2n?12?0.
所以当n?6时,cn?1?cn.因此当n?6时,cn?c6?于是当n?6时,
n(n?2)22?34?1.
?1.
1n.
综上所述,当n?6时,Sn?2?11.(陕西卷22).(本小题满分14分) 已知数列{an}的首项a1?35,an?1?3an2an?12?. ,n?1,,(Ⅰ)求{an}的通项公式;
11?x?2?2?; ?x??,n?1,,2n(1?x)?3?1(Ⅱ)证明:对任意的x?0,an≥?(Ⅲ)证明:a1?a2???an?n2n?1.
解法一:(Ⅰ)?an?1?3an2an?1,?1an?1?23?13an,?1an?1?1??1?1?1??, 3?an?又
1an1an?1??1?21?1?是以为首项,为公比的等比数列. ,??333?an?22?n?1?n,?an?n.
3333?212??1?3n(Ⅱ)由(Ⅰ)知an?3nn3?2?0,
延津县高级中学 z.hxw.2008@163.com
郑学伟 第 15 页 2013-4-15
11?x1??2??x? 2?n(1?x)?3??2??1?1?x? 2?n(1?x)?3?11?1?x??11?x1??1??(1?x)? 2?(1?x)?an?1??12 ??2an(1?x)1?x21?1????an??an≤an,?原不等式成立. ?an?1?x?(Ⅲ)由(Ⅱ)知,对任意的x?0,有
a1?a2???an≥11?x?1111?2?2??2???x???x?????x?????? 2222n(1?x)?31?x(1?x)?3?1?x(1?x)?3??1?n1?x?22?2??????nx?. 2?2n(1?x)?333?12?1?1??n?1?222?3?1?3?1???1?n?, ?取x???2???n??1?n?333?n?3??n?1??3??n1?1?1??1?n?n?3?n2则a1?a2???an≥?n?1?13n?n2n?1.
?原不等式成立.
解法二:(Ⅰ)同解法一. (Ⅱ)设f(x)?11?x??2??x??, 2n(1?x)?3?1?2??2?2?(1?x)??n?x??2(1?x)2?n?x?1?3??3?则f?(x)?? ??222(1?x)(1?x)(1?x)?x?0,
?当x?23n时,f?(x)?0;当x?23n时,f?(x)?0,
延津县高级中学 z.hxw.2008@163.com
郑学伟 第 16 页 2013-4-15
1?2?时,取得最大值f??an. f(x)?n?n233??1?n3?原不等式成立.
?当x?2(Ⅲ)同解法一.
12.(重庆卷22)(本小题满分12分,(Ⅰ)小问5分,(Ⅱ)小问7分.)
3 设各项均为正数的数列{an}满足a1?2,an?aa2?1aa?2(n?N*).
(Ⅰ)若a2?14,求a3,a4,并猜想a2cos的值(不需证明);
(Ⅱ)记bn?a3a2???an(n?N*),若bn?22对n≥2恒成立,求a2的值及数列{bn}的通项公式.
解:(Ⅰ)因a1?2,a2?2?2,故
?32324
a3?a1a2?2,?2.0
a4?a2a3??8 由此有a1?2(?2),a2?2(?2),a3?2(?2),a4?2(?2),故猜想an的通项为 an?2(?2)n?1223(n?N).
Sn* (Ⅱ)令xn?log2an,Sn表示xn的前n项和,则bn?2. 由题设知x1=1且
xn?32xn?1?xn?2(n?N); ①
*
Sn?x1?x2???xn? 因②式对n=2成立,有 x2?123232(n?2). ②
?x1?x2,又x1?1得
. ③ 12.假设x2?3212.
12(xn?1?2xn). 12 下用反证法证明:x2? 由①得xn?2?2xn?1?(xn?2?xn?1)? 因此数列xn?1?2xn是首项为x2?2,公比为 xn?1?12xn?(x2?1)1n?1*的等比数列.故
22(n?N). ④
延津县高级中学 z.hxw.2008@163.com
相关推荐: