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aaaa;mnpq(2)数列a2n,an,an仍为等差数列,Sn,S2nSn,S3nS2n⋯⋯仍为等差数12212;列,公差为nd(3)若三个成等差数列,可设为ad,a,ad(4)若a,b是等差数列,且前n项和分别为nnaSm2m1S,T,则nnbTm2m1(5)a为等差数列n2Sanbn(a,b为常数,是关于n的常数项为0的二n次函数)S的最值可求二次函数n2Sanbn的最值;或者求出an中的正、负分界n项,即:当anna10,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,1蓝天教育辅导中心独家经典讲义S2n(2n1)an(an为中间项),1S奇Sa,:anan1q(q为常数,q0),:x、G、y成等比数列G2xy,(q1)1前n项和:nSa1qn11q(q1)(要注意!)性质:a是等比数列n(1)若mnpq,则a·aa·amnpq(2)S,SS,SS⋯⋯仍为等比数列,:由S求na时应注意什么?nn1时,aS;11n2时,(1)求差(商)法如:数列111a,a1a2⋯⋯a2n5,求ann2nn2221解n1时,a1,∴a114①2152111n2时,a1a2⋯⋯a12n15②n2n1222①—②得:1na2,∴n2n1a2,∴n14(n1)ann12(n2)[练习]数列5a满足SS1a1,a14,求nnnn3an注意到aSS,代入得n1n1nSn14又S;nnS14,∴Sn是等比数列,S4n2蓝天教育辅导中心独家经典讲义n2时,naSS1⋯⋯3·4nnn1(2)叠乘法如:数列a中,nana13,,求n1an1nan解aaa12n123n·⋯⋯·⋯⋯,∴aaa23n12n1a1nan1又a13,∴an3n.(3)等差型递推公式由aa1f(n),a1a0,求an,用迭加法nnaaf21(2)n2时,aaf32⋯⋯⋯⋯(3)两边相加得aa1f(2)f(3)⋯⋯f(n)naaf(n)nn1∴aa0f(2)f(3)⋯⋯f(n)n[练习]数列a中,nn1a,aan,求an(11312nn11312nnan12n31)(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111axcaxacacxnnnn令(c1)xd,∴xdc1,∴ancd1是首项为da,c为公比的等比数列1c1∴dddd·n,∴1n1c1c1c1c1(5)倒数法如:2a,,求na1a1n1a2nan由已知得:1a211na2a2an1nn,∴111aan1n2∴1an为等差数列,1a11,公差为12,∴1111n1·n1,a22n3蓝天教育辅导中心独家经典讲义∴ann21(附:S1(n1)an公式法、利用1(2)SSn、累加法、(n)、待定系数法、对数变换法、迭代法、数学归纳法、n1n换元法)(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解:由11111a·aaaddaakk1kkkk1d0∴nn**********⋯⋯kakakkdakakdaaaaanan1**********daa1n1[练习]求和:1111⋯⋯12123123⋯⋯naS⋯⋯⋯⋯,nn2n11(2)错位相减法若a为等差数列,nb为等比数列,求数列nab(差比数列)前n项和,可由nnSqS,求Sn,:23n1S12x3x4x⋯⋯nx①n234n1nx·Sx2x3x4x⋯⋯n1xnx②n①—②2n1n1xS1xx⋯⋯xnxn4蓝天教育辅导中心独家经典讲义x1时,Sn11nnxnx21xx,x1时,S123⋯⋯nnnn21(3)倒序相加法把数列的各项顺序倒写,⋯⋯aan12n1nSaa⋯⋯aannn121相加2Snaanaan⋯aan⋯1211[练习]已知f(x)12x2x,则111f(1)f(2)ff(3)f