2022年数列解题技巧归纳总结-打印3 .pdf
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1、。-可编辑修改-等差数列前n项和的最值问题:1、若等差数列na的首项10a,公差0d,则前n项和nS有最大值。()若已知通项na,则nS最大100nnaa;()若已知2nSpnqn,则当n取最靠近2qp的非零自然数时nS最大;2、若等差数列na的首项10a,公差0d,则前n项和nS有最小值()若已知通项na,则nS最小100nnaa;()若已知2nSpnqn,则当n取最靠近2qp的非零自然数时nS最小;数列通项的求法:公式法:等差数列通项公式;等比数列通项公式。已知nS(即12()naaaf n)求na,用作差法:11,(1),(2)nnnSnaSSn。已知12()na aaf n求na,用作
2、商法:(1),(1)(),(2)(1)nfnf nanf n。已知条件中既有nS还有na,有时先求nS,再求na;有时也可直接求na。若1()nnaaf n求na用累加法:11221()()()nnnnnaaaaaaa1a(2)n。已知1()nnaf na求na,用累乘法:121121nnnnnaaaaaaaa(2)n。已知递推关系求na,用构造法(构造等差、等比数列)。特别地,(1)形如1nnakab、1nnnakab(,k b为常数)的递推数列都可以用待定系数法转化为公比为k的等比数列 后,再求na;形如1nnnakak的递推数列都可以除以nk得到一个等差数列后,再求na。(2)形如11n
3、nnaakab的递推数列都可以用倒数法求通项。(3)形如1knnaa的递推数列都可以用对数法求通项。(7)(理科)数学归纳法。-可编辑修改-(8)当遇到qaadaannnn1111或时,分奇数项偶数项讨论,结果可能是分段一、典型题的技巧解法1、求通项公式(1)观察法。(2)由递推公式求通项。对于由递推公式所确定的数列的求解,通常可通过对递推公式的变换转化成等差数列或等比数列问题。(1)递推式为 an+1=an+d 及 an+1=qan(d,q 为常数)例 1、已知 an 满足 an+1=an+2,而且 a1=1。求 an。例 1、解an+1-an=2 为常数an是首项为1,公差为2 的等差数列
4、an=1+2(n-1)即 an=2n-1 例 2、已知na满足112nnaa,而12a,求na=?(2)递推式为 an+1=an+f(n)例 3、已知na中112a,12141nnaan,求na.解:由已知可知)12)(12(11nnaann)121121(21nn令 n=1,2,(n-1),代入得(n-1)个等式累加,即(a2-a1)+(a3-a2)+(an-an-1)2434)1211(211nnnaan说明只要和f(1)+f(2)+f(n-1)是可求的,就可以由an+1=an+f(n)以 n=1,2,(n-1)代入,可得n-1 个等式累加而求an。(3)递推式为 an+1=pan+q(p
5、,q 为常数)例 4、na中,11a,对于 n1(n N)有132nnaa,求na.解法一:由已知递推式得an+1=3an+2,an=3an-1+2。两式相减:an+1-an=3(an-an-1)因此数列 an+1-an是公比为3 的等比数列,其首项为a2-a1=(3 1+2)-1=4 an+1-an=43n-1an+1=3an+23an+2-an=43n-1 即 an=23n-1-1 解法二:上法得 an+1-an是公比为3 的等比数列,于是有:a2-a1=4,a3-a2=4 3,a4-a3=4 32,an-an-1=4 3n-2,把 n-1 个等式累加得:an=23n-1-1 文档编码:C
6、U4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6
7、A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:C
8、U4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6
9、A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:C
10、U4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6
11、A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:C
12、U4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7文档编码:CU4T5A5Z7X4 HD7W1H5H6A5 ZI5A2H3V4Z7。-可编辑修改-(4)递推式为 an+1=p an+q n(p,q 为常数))(3211nnnnbbbb由上题的解法,得:nnb)32(23nnnnnba)31(2)21(32(5)递推式为21nnnapaqa思路:设21nnnapaqa,可以变形为:211()nnnnaaaa,想于是 an+1-an是公比为的等比数列,就转化为前面的类型。求na。(6)递推式为 Sn与
13、 an的关系式文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P
14、8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T
15、1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P
16、8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T
17、1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P
18、8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T
19、1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3。-可编辑修改-关系;(2)试用 n表示 an。)2121()(1211nnnnnnaaSS11121nnnnaaannnaa21211上式两边同乘以2n+1得 2n+1an+1=2nan+2 则2nan 是公差为 2 的等差数列。2nan=2+(n-1)2=2n 2数列求和问题的方法(1)、应用公式法等差、等比数列可直接利用等差、等比数列的前
20、n项和公式求和,另外记住以下公式对求和来说是有益的。13 5(2n-1)=n2【例 8】求数列 1,(3+5),(7+9+10),(13+15+17+19),前 n 项的和。解本题实际是求各奇数的和,在数列的前n 项中,共有1+2+n=)1(21nn个奇数,最后一个奇数为:1+21n(n+1)-12=n2+n-1 因此所求数列的前n 项的和为(2)、分解转化法对通项进行分解、组合,转化为等差数列或等比数列求和。【例 9】求和 S=1(n2-1)+2 (n2-22)+3(n2-32)+n(n2-n2)解 S=n2(1+2+3+n)-(13+23+33+n3)文档编码:CC5C4W5P8H4 HL
21、1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O
22、3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL
23、1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O
24、3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL
25、1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O3文档编码:CC5C4W5P8H4 HL1T5H4Z7R7 ZK8T1Q4F3O
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