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2302345:专题二 第四讲 数学思想方法与答题模板建构(理)_图文

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第4讲
数学思 想方法 活用数学思想 追求高效解题

与答题
模板建 构(理) 巧用答题模板 建立答题规范

1.分类讨论思想
分类讨论思想就是将一个复杂的数学问题分解成若干个简 单的基础性问题,通过对基础性问题的解答,解决原问题 的思维策略.实质上,分类讨论是“化整为零,各个击破, 再积零为整”的数学策略,分类讨论可以优化解题思路, 降低问题难度.分类的原则是:(1)分类的对象确定,标准 统一;(2)不重复,不遗漏;(3)分层次,不越级讨论.

在等比数列求和中经常对公比q进行分类,而有的数列通
项公式以分段函数给出,或以(-1)n形式给出的,要分类 求解.求含参数极限有时也要分类讨论.

[例1] (2011· 四川高考)已知{an}是以a为首项,q为公比的等 比数列,Sn为它的前n项和. (1)当S1,S3,S4成等差数列时,求q的值; (2)当Sm,Sn,Sl成等差数列时,求证:对任意自然数k,am+
k,an+k,al+k也成等差数列.

[解]

(1)由已知,an=aqn 1,因此


S1=a,S3=a(1+q+q2),S4=a(1+q+q2+q3). 当S1,S3,S4成等差数列时,S4-S3=S3-S1, 可得aq3=aq+aq2, 1± 5 化简得q -q-1=0.解得q= . 2
2

(2)若q=1,则{an}的每项an=a, 此时am+k,an+k,al+k显然构成等差数列. 若q≠1,由Sm,Sn,Sl构成等差数列可得Sm+Sl=2Sn, a?qm-1? a?ql-1? 2a?qn-1? 即 + = . q- 1 q-1 q-1 整理得qm+ql=2qn. 因此,am+k+al+k=aqk-1(qm+ql)=2aqn+k-1=2an+k, 所以,am+k,an+k,al+k也成等差数列.

2.在等差(比)数列的通项公式和前n项和公式中共有5个量

a1、d(或q)、n、an及Sn,这5个量中知道其中任意3个量
的值,就可以通过运用方程思想,解方程(或方程组)求 出另外2个量的值.

[例2] (2011· 江西高考)已知两个等比数列{an},{bn},满 足a1=a(a>0),b1-a1=1,b2-a2=2,b3-a3=3. (1)若a=1,求数列{an}的通项公式;

(2)若数列{an}唯一,求a的值.

[解]

(1)设数列{an}的公比为q,

则b1=1+a=2,b2=2+aq=2+q,b3=3+aq2=3+q2, 由b1,b2,b3成等比数列得(2+q)2=2(3+q2). 即q2-4q+2=0,解得q1=2+ 2,q2=2- 2. 所以数列{an}的通项公式为an=(2+ 2)n-1或an=(2- 2)n-1.

(2)设数列{an}的公比为q, 则由(2+aq)2=(1+a)(3+aq2), 得aq2-4aq+3a-1=0(*), 由a>0得Δ=4a2+4a>0, 故方程(*)有两个不同的实根. 1 由数列{an}唯一,知方程(*)必有一根为0,代入(*)得a= . 3

[命题角度分析] 近两年来,高考关于数列方面的命题主要有以下三个方面:

(1)数列本身的有关知识,其中有等差数列与等比数列的概念、
性质、通项公式及求和公式; (2)数列与其他知识的结合,其中有数列与函数、方程、不等式、 三角、几何的结合; (3)数列的应用问题,其中主要是以增长率问题为主.

[答题模板构建]

[例 3]

(本小题满分 12 分) (2011· 浙江高考)已知公差不为 0 的等差数
1 2 4

1 1 1 列{an}的首项 a1 为 a(a∈R),且a ,a ,a 成等比数列. (1)求数列{an}的通项公式; 1 1 1 1 1 (2)对 n∈N*,试比较a + a + a +?+ a 与a 的大小. 2 1 22 23 2n

[解]

(1)设等差数列{an}的公差为d,

1 1 1 由题意可知( )2= · , a2 a1 a4 即(a1+d)2=a1(a1+3d),从而a1d=d2.┄┄┄┄┄┄┄┄┄┄┄(2分) 因为d≠0.所以d=a1=a.┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄(5分) 故通项公式an=na.┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄(6分)

?第一步 列出含有a1和d的方程; ?第二步 推出a1=d=a; ?第三步 写出an.

1 1 1 (2)记 Tn= + a +?+ a ,因为 a2n=2na,┄┄┄┄┄┄(7 分) a2 22 2n 1 1n ?1-? ? ? 2 11 1 1 12 所以 Tn=a( + 2+?+ n)=a· 2 2 2 1 1- 2 1 1 =a[1-( )n].┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄(10 分) 2 1 1 从而,当 a>0 时,Tn< ;当 a<0 时,Tn> .┄┄┄┄(12 分) a1 a1

?第一步 写出Tn的表达式; ?第二步 利用求和公式求Tn; ?第三步 分情况下结论.

[例4] (2011· 浙江联考)数列{an}满足an+1+an=4n-
3(n∈N*). (1)若{an}是等差数列,求其通项公式; (2)若{an}满足a1=2,Sn为{an}的前n项和,求S2n+1.

[解] 由题意得 an+1+an=4n-3,① an+2+an+1=4n+1,② 由②-①得 an+2-an=4, (1)∵{an}是等差数列,设公差为 d,∴d=2. 1 又∵a1+a2=1,∴a1+a1+d=1.∴a1=-2. 5 ∴an=2n-2.

(2)∵a1=2,a1+a2=1,∴a2=-1. 又∵an+2-an=4, ∴数列的奇数项与偶数项分别成等差数列,公差均为 4. S2n+1=(a1+a3+?+a2n+1)+(a2+a4+?+a2n) ?n+1?n n?n-1? =(n+1)×2+ 2 ×4+n×(-1)+ 2 ×4 =4n2+n+2.

[点评] 本题考查数列的递推公式、等差数列的定义、通项

公式、前n项和公式的应用.由递推公式得到an+2-an=
4(n∈N*)是解决本题的关键.

[例 5]

(2011?大连联考)已知数列{an}各项均为正数, 其前 n 项和为

Sn,且满足 4Sn=(an+1)2. (1)求{an}的通项公式; 1 (2)设 bn= ,数列{bn}的前 n 项和为 Tn,求 Tn 的最小值. an · an+1

[解]

(1)因为(an+1)2=4Sn,所以

?an+1?2 ?an+1+1?2 Sn= ,Sn+1= . 4 4 所以Sn+1-Sn=an+1 ?an+1+1?2-?an+1?2 = . 4
2 即4an+1=a2 n+1-an+2an+1-2an,

∴2(an+1+an)=(an+1+an)· (an+1-an). 因为an+1+an≠0,所以an+1-an=2. 即{an}为公差等于2的等差数列. 由(a1+1)2=4a1,解得a1=1,所以an=2n-1.

1 (2)由(1)知bn= ?2n-1??2n+1? 1 ? 1? ? 1 ? = ?2n-1-2n+1?, 2? ? ∴Tn=b1+b2+?+bn 1 1 1 1 1 ? 1? ? ? = ?1-3+3-5+?+2n-1-2n+1? 2? ? 1 ? 1? ? ? = ?1-2n+1? 2? ? 1 1 = - . 2 2?2n+1?

?1 ? 1 1 1 ? ∵Tn+1-Tn= - -?2-2?2n+1?? ? 2 2?2n+3? ? ?

1 1 = - 2?2n+1? 2?2n+3? 1 = >0, ?2n+1??2n+3? ∴Tn+1>Tn.∴数列{Tn}为递增数列, 1 1 1 ∴Tn的最小值为T1= - = . 2 6 3

[点评] 本题考查数列的通项与前n项和的关系,以及等差数

列的定义,单调数列的判断等内容.转化条件4Sn=(an+1)2
为解题的关键.


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