文档介绍:数列公式总结数列基础知识点和方法归纳 :an?1?an?d,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?na 2 1? n?n?1? d2 性质:?an?是等差数列若m?n?p?q,则am?an?ap?aq; 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; 若三个成等差数列,可设为a?d,a,a?d若an,bn是等差数列,且前n项和分别为Sn,Tn,则 amS2m?1 ? bmT2m?1 ?an?为等差数列?Sn?an2?bn 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?可得Sn达到最大值时的n值. a?0?(转载于:写论文网:数列公式总结)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? an .an?1 ,有项数为奇数2n?1的等差数列?an? 1 S2n?1?(2n?1)an(an为中间项),S奇?S奇偶?an, SS? n偶 n?1 .: an?1 a?q,an?a1qn?1n .等比中项:x、G、y成等比数列?G2? xy,或G? ?na1(q前n项和:S? ?1)n??? a1?1?qn??1?q (q?1) 性质:?an?是等比数列若m?n?p?q,则am·an?ap·aq Snn,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. :数列?a12?11 n?,a122a2?……?2 nan?2n?5,求an 解n?1时,1 2a1?2?1?5,∴a1?14n?2时,12a?11 122a2?……?2 n?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?a5 n?满足Sn?Sn?1?3 an?1,a1?4,求an 注意到aSn?1 n?1?Sn?1?Sn,代入得 S?4又S1?4,∴?Sn?是等比数列,n ; 2 ①② Sn?4n n?2时,an?Sn?Sn?1?……?3·4n?1 叠乘法 an如:数列?an?中,a1?3n?1?,求an ann?1 解 3aa1a2a312n?1 ,∴n?又a1?3,∴an?……n?…… ?123n 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等比型递推公式 n?1 ?an?1?n?2?,求an an?can?1?d 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? 倒数法如:a1?1,an?1? 2an ,求anan?2 由已知得: a??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、 an?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? 1 k?1akak?1 n 解:由 n 111?11? ??????d?0? ak·ak?1akak?dd?akak?1? n ?111?11?1??11??11?1?? ?????????……??∴?????????? ak?1?d??a1a2??a2a3?k?1