高中数学数列知识点总结.doc

文档介绍：高中数学数列知识点总结篇一:高中数学数列知识点(经典):an?1?an?d(d为常数),an?a1??n?1?d等差中项:x,A,y成等差数列?2A?x?y前n项和Sn??a1?an?n?na21?n?n?1?d2性质:?an?是等差数列(1)若m?n?p?q,则am?an?ap?aq;(2)数列?a2n?1??,a2n??,a2n?1?仍为等差数列,Sn,S2n?Sn,S3n?S2n……仍为等差数列,公差为n2d;(3)若三个成等差数列,可设为a?d,a,a?d(4)若an,bn是等差数列,且前n项和分别为Sn,Tn,则amS2m?1?bmT2m?1(5)?an?为等差数列?Sn?an2?bn(a,b为常数,是关于n的常数项为0的二次函数)Sn的最值可求二次函数Sn?an2?bn的最值;或者求出?an?中的正、负分界项,?an?0即:当a1?0,d?0,解不等式组??0?n?1?a?0当a1?0,d?0,由?n可得Sn达到最小值时的n值.?an?1?0(6)项数为偶数2n的等差数列?an?,有S2n?n(a1?a2n)?n(a2?a2n?1)???n(an?an?1)(an,an?1为中间两项)S偶?S奇?nd,S奇S偶??1,有(7)项数为奇数2n?1的等差数列?an?1S2n?1?(2n?1)an(an为中间项),S奇?SS奇偶?an,S?nn?:an?1?q(q为常数,q?0),an?1an?:x、G、y成等比数列?G2?xy,或G??na1(q?1)前n项和:S?n???a1?1?qn?(要注意!)?1?q(q?1)性质:?an?是等比数列(1)若m?n?p?q,则am·an?ap·aq(2)Sn,S2n?Sn,S3n?S2n……仍为等比数列,:由Sn求an时应注意什么?n?1时,a1?S1;n?2时,an?Sn?Sn?(1)求差(商)法如:数列?a12?11n?,a122a2?……?2nan?2n?5,求an解n?1时,12a1?2?1?5,∴a1?14n?2时,12a?11122a2?……?2n?1an?1?2n?1?5①—②得:1n?1?14(n?1)2nan?2,∴an?2,∴an???2n?1(n?2)[练习]数列?a5n?满足Sn?Sn?1?3an?1,a1?4,求an注意到aSn?1n?1?Sn?1?Sn,代入得S?4又S1?4,∴?Sn?是等比数列,n;2①②Sn?4nn?2时,an?Sn?Sn?1?……?3·4n?1(2)叠乘法an如:数列?an?中,a1?3n?1?,求anann?1解3aa1a2a312n?1,∴n?又a1?3,∴an?……n?……?123n(3)等差型递推公式由an?an?1?f(n),a1?a0,求an,用迭加法?a3?a2?f(3)??n?2时,?两边相加得an?a1?f(2)?f(3)?……?f(n)…………?an?an?1?f(n)??a2?a1?f(2)∴an?a0?f(2)?f(3)?……?f(n)[练习]数列?an?中,a1?1,an?3(4)等比型递推公式n?1?an?1?n?2?,求an(an?1n3?1??2)an?can?1?d(c、d为常数,c?0,c?1,d?0)可转化为等比数列,设an?x?c?an?1?x??an?can?1??c?1?x令(c?1)x?d,∴x?ddd??,c为公比的等比数列,∴?an??是首项为a1?c?1c?1c?1??∴an?dd?n?1d?n?1d??,∴??a1?·ca?a?c?n??1?c?1?c?1?c?1?c?1?(5)倒数法如:a1?1,an?1?2an,求anan?2由已知得:a?2111111?n??,∴??an?12an2anan?1an2?1?11111·??n?1?,∴??为等差数列,?1,公差为,∴?1??n?1?2a1an22?an?3∴an?(附:2n?1公式法、利用an??S1(n?1)Sn?Sn?1(n?2)、累加法、?1?pan?q或an?1?pan?f(n)、待定系数法、对数变换法、迭代法、数学归纳法、换元法)(1)裂项法把数列各项拆成两项或多项之和,:?an?是公差为d的等差数列,求?1k?1akak?1n解:由n111?11???????d?0?ak·ak?1akak?dd?akak?1?n?111?11?1??11??11?1???????????……??∴??????????ak?1?d?