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1、类型一：1()nnaaf n（fn可以求和）解决方法累加法1.已知数列na，1a=2，1na=na+3n+2，求na。2.已知数列an满足3a132aa1nn1n，求数列an的通项公式。答案：1.(31)2nnna2.2.31nnan类型二：1()nnaf na（()f n可以求积）解决方法累乘法1.已知数列na满足321a，nnanna11，求na。2.已知31a，nnanna23131)1(n，求na。答案：1.23nan2.631nan类型三：1(nnaAaB 其中A,B为常数A0,1）解决方法待定系数法1.在数列na中，11a，当2n时，有132nnaa，求数列na的通项公式。答案：设
2、13nnatat，则132nnaat1t，于是1131nnaa1na是以112a为首项，以 3 为公比的等比数列。12 31nna2.在数列 an中，,13,111nnaaa求na.答案：1132nna3.设在数列na中，11a，112122nnaann求数列na的通项公式。答案：设nnbaAnb1112nnaAnBaA nB展开后比较得204261022AAABB这时11462nnnnbbann2 且bnb是以 3 为首项，以12为公比的等比数列1132nnb即113462nnan，113462nnan4.在数列na中，12a，11222nnnaan求数列na的通项公式。答案：11222nn
3、naan1122nnnaa，两边同除以2n得11222nnnnaa2nna是以12a=1 为首项，2 为公差的等差数列。112212nnann即221nnan类型四：()nnSf a解决方法11(1)(2)nnnsnassn1.已知数列na前 n 项和2214nnnaS.1求1na与na的关系；（2）求通项公式na.答案：1文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9
4、J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4
5、N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M
6、5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2
7、Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10
8、K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C
9、3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2
10、Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M511n时，11142asa，得11a；22n时，1123114422nnnnnnnassaa；得11122nnnaa。（2）在上式中两边同乘以12n得11222nnnnaa；2nna数列是以1122a为首项，2 为公差的等差数列；22222nnann；得12nnna。2.已知在正整数数列na中，前n项和nS满足21(2)8nnSa，求数列na的通项公式.答案：42nan3.数列na前 n 项和2214nnnaS.（1）求1na与na的关系；（2）求通项公式na.答案：（1）由2214nnnaS得：111214nnnaS于是)2121
11、()(1211nnnnnnaaSS，所以11121nnnnaaannnaa21211.（2）上式两边同乘以12n得：22211nnnnaa由1214121111aaSa.于是数列nna2是以 2 为首项，2 为公差的等差数列，所以nnann2)1(22212nnna类型五：迭代类型六：裂项相消如)2(1nnan求 Sn。常用的裂项有111)1(1nnnn；)211(21)2(1nnnn；文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F
12、10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U
13、4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8
14、P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B
15、3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V
16、4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:C
17、Z8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD
18、8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5)2)(1(1)1(121)2)(1(1nnnnnnn1.（湖北卷）已知二次函数()yf x的图像经过坐标原点，其导函数为()62fxx，数列na的前 n 项和为nS，点(,)()nn SnN均在函数()yf x的图像上。（）、求数列na的通项公式；（）、设11nnnba a，nT是数列nb的前 n 项和，求使得20nmT对所有nN都成立的最小正整数m；解：（）设这二次函数f(x)ax2+bx(a 0),则 f(x)=2ax+b,由于 f(x)=6x2,得a=3,b=2
19、,所以f(x)3x22x.又因为点(,)()nn SnN均在函数()yf x的图像上，所以nS3n22n.当 n2 时，anSnSn1（3n22n）)1(2)132nn（6n5.当 n1 时，a1S13122615，所以，an6n5（nN）（）由（）得知13nnnaab5)1(6)56(3nn)161561(21nn，故 Tnniib121)161561(.)13171()711(nn21（1161n）.因此，要使21（1161n）20m（nN）成立的 m,必须且仅须满足2120m，即 m 10，所以满足要求的最小正整数m 为 10.2.求数列,11,321,211nn的前 n 项和.解：设n
20、nnnan111则11321211nnSn)1()23()12(nn11n文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6
21、HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5
22、ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档
23、编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J
24、6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N
25、5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5
26、文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5类型七：错位相减法（2014?濮阳二模）设an是等差数列，bn是各项都为正数的等比数列，且a1=b1=1，a3+b5=21，a5+b3=13（）求 an、bn的通项公式；（）求数列的前 n
27、项和 Sn解答：解：（）设 an的公差为d，bn的公比为q，则依题意有q0 且解得 d=2，q=2所以 an=1+（n 1）d=2n1，bn=qn1=2n1（），得，=文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z
28、2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F1
29、0K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4
30、C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P
31、2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3
32、F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5文档编码:CZ8P2Z2Q9J6 HD8B3F10K4N5 ZD1V4U4C3M5