文档介绍：:aad(d为常数),n1naand11n等差中项:x,A,y成等差数列2Axy前n项和aannn11nSnadn122性质:a是等差数列n(1)若mnpq,则amanapaq;2;(2)数列a2n1,a2n,a2n1仍为等差数列,Sn,S2nSn,S3nS2n⋯⋯仍为等差数列,公差为nd(3)若三个成等差数列,可设为ad,a,adaSm2m1(4)若an,bn是等差数列,且前n项和分别为Sn,Tn,则bTm2m1(5)an为等差数列2Sanbn(a,b为常数,是关于n的常数项为0的二次函数)nS的最值可求二次函数n2Sanbn的最值;或者求出a中的正、负分界项,nn即:当ana10,d0,,d0,(6)项数为偶数2n的等差数列an有,S2nn(a1a2n)n(a2a2n1)n(anan1)(an,an1为中间两项)S偶Snd,(7)项数为奇数2n1的等差数列an有,S2n(nanan为中间项,21)()1S奇S偶a,:n1anq(q为常数,q0),:x、G、y成等比数列2Gxy,(q1)1前n项和:nSa1qn11q(q1)(要注意!)性质:a是等比数列n(1)若mnpq,则am·anap·aq(2)S,SS,SS⋯⋯仍为等比数列,:由S求an时应注意什么?nn1时,a1S1;n时,(1)求差(商)法如:数列111a,a1a2⋯⋯an,求an25n2nn222(2)叠乘法如:数列a中,nana13,,求ann1an1n(3)等差型递推公式由aa1f(n),a1a0,求an,用迭加法nn2数列基础知识点和方法归纳n1a,aan,求an(11312nnan12n31[练习]数列an中,)(4)等比型递推公式acad(c、d为常数,c0,c1,d0)nn1可转化为等比数列,设axca1xaca1c1xnnnn令(c1)xd,∴xcd,∴1ancd1d是首项为a1,c为公比的等比数列c1∴ddn1aa·c,∴(5)倒数法如:2ana1,a,求an1n1a2n附:S1(n1)an公式法、利用SS1(n2)nn、累加法、(n)、待定系数法、对数变换法、迭代法、数学归纳法、换元法nn)(1)裂项法把数列各项拆成两项或多项之和,:a是公差为d的等差数列,求nn1kakak11(2)错位相减法若a为等差数列,bn为等比数列,求数列anbn(差比数列)前n项和,可由SnqSn,求Sn,:23n1S12x3x4x⋯⋯nx①n234n1nx·Sx2x3x4x⋯⋯n1xnx②n①—②2n1n1xS1xx⋯⋯xnxnx1时,Sn11nnxnx21xx,x1时,S123⋯⋯nnnn21(3)倒序相加法把数列的各项顺序倒写,⋯⋯n12n