高考数列知识点精华
总结
初级经济法重点总结下载党员个人总结TXt高中句型全总结.doc高中句型全总结.doc理论力学知识点总结pdf
数列
1?重要公式:1+2+…+n=n(n+1); 2
12221+2+…+n=n(n+1)(2n+1); 6
13332221+2+…+n=(1+2+…+n)=n(n+1); 4
43. 等差数列的定义与性质
,, 定义:a,a,d(d为常数),a,a,n,1dn,1nn1
等差中项:x,A,y成等差数列,2A,x,y
aan,,,nn1,,,1n前项和nSnad ,,, 1n22
,,性质:a是等差数列n
(1)若m,n,p,q,则a,a,a,a;mnpq
,,,,,,(2)数列a,a,ka,b仍为等差数列;2n,12nn
S,S,S,S,S……仍为等差数列;n2nn3n2n
(3)若三个数成等差数列,可设为a,d,a,a,d;
aSmm2,1abSTn (4)若,是等差数列,为前项和,则,;nnnnbTm2m,1
2 ()为等差数列(,为常数,是关于的常数项为5aSanbnabn,,,,,nn
0的二次函数)
2 SSanbna的最值可求二次函数的最值;或者求出中的正、负分界,,,,nnn
项,即:
,a,0n,,当a0,d0,解不等式组可得S达到最大值时的n值。 ,1n,0a,1,n
,a,0n,,当a0,d0,由可得S达到最小值时的n值。 ,1n,0a,1,n
,, 如:等差数列a,S,18,a,a,a,3,S,1,则n, nnnn,1n,23
(由a,a,a,3,3a,3,?a,1nn,1n,2n,1n,1
aa,,,113又?331,?S,,a,a, 32223
,,1,,,1n,,3,,,,,,?aanaan,,1n2n,1 ?,,,,18Sn222
?,n27)
44. 等比数列的定义与性质
a,1nn,1 定义:,q(q为常数,q,0),a,aq1nan
2 等比中项:x、G、y成等比数列,G,xy,或G,,xy
naq,,(1)1,nS, 前n项和:(要注意!),aq,,,1n1q,(1),,1q,
性质:是等比数列a,,n
(1)若m,n,p,q,则a?a,a?amnpq
(2)S,S,S,S,S……仍为等比数列n2nn3n2n
45.由求时应注意什么,Sann
(n,1时,a,S,n,2时,a,S,S)11nnn,1
46. 你熟悉求数列通项公式的常用方法吗,
例如:(1)求差(商)法
111,,如:a满足a,a,……,a,2n,5,1, n12n2n222
1 解: naa,,,,,1时,,?21514112
111n,2时,a,a,……,a,2n,1,5,2, 12n,12,1n222
1,1,,,2,得:a,2 nn2
n,1 ?a,2n
n,,14(1)a ?,,nn,12(n2),,
,练习,
5,, 数列a满足S,S,a,a,4,求a nnn,1n,11n3
Sn,1aSS (注意到,,代入得:,4nnn,1,1Sn
n ,,又S,4,?S是等比数列,S,41nn
n,1 n,2时,a,S,S,……,3?4nnn,1
(2)叠乘法
ann,1 ,,,,例如:数列a中,a3,,求ann1,1ann
aaaan,121123nn 解: ,,?……?……,?aaanan2312,11n
3 又,?aa,,31nn
(3)等差型递推公式
由a,a,f(n),a,a,求a,用迭加法nn,110nnaaf,,,,2时,(2)21,aaf,,(3),32 两边相加,得:,…………,
,a,a,fn()nn,1,
aafffn,,,,,()()()23……n1
?a,a,f(2),f(3),……,f(n)n0
,练习,
n,1 ,,,, 数列a,a,1,a,3,an,2,求an1nn,1n
1n ()a,,31,,n2
(4)等比型递推公式
,, a,ca,dc、d为常数,c,0,c,1,d,0nn,1
,, 可转化为等比数列,设a,x,ca,xnn,1
,,,a,ca,c,1xnn,1
dcxdx令(,1),,?, c,1
dd,,?是首项为,为公比的等比数列a,a,c ,,n1c,11c,,,
,,dd,1n ,,,,,?aa?c1n,,,,11cc,,
dd,,,1naac ,,?,,,1n,,c,1c,1,,
,练习,
,,数列a满足a,9,3a,a,4,求an1n,1nn
n,1,,4 ,, (,8,,1)an,,3,,
(5)倒数法
a2n ,,例如:a1,a,求a1n,1n,2an
a,2111n ,,,由已知得:aaa22n,1nn
111 ?,,aa2nn,1
,,111为等差数列,1,公差为 ?,,,aa2n1,,
111,,,,?,1,n,1?,n,1 22an
2a?, nn,1
47. 你熟悉求数列前n项和的常用方法吗,
例如:(1)裂项法:把数列各项拆成两项或多项之和,使之出现成对互为相反数的项。
n1如:是公差为的等差数列,求ad ,,,naak,1kk,1
,,11111,, 解: 由,,,d,0,,,,,,?,aaaaddaakk,1kkkk,1,,
nn,,1111,, ,,?,,,,aadaa,1,1,1,1kk,,kkkk
,,,,,,,,1111111,,,,,,,,,,,,,……,,,,,,,,daaaaaann,,,,,,1223,1,,
,,111,,,,,,daa1n,1,,
,练习,
111求和:1,,,……, 1,21,2,31,2,3,……,n
1 (…………,)aS,,,,2nnn,1
(2)错位相减法:
若为等差数列,为等比数列,求数列(差比数列)前项ababn,,,,,,nnnn
和,可由求,其中为的公比。SqSSqb,,,nnnn
23n,1 如:S,1,2x,3x,4x,……,nx,1,n
234n,1n ,,x?S,x,2x,3x,4x,……,n,1x,nx,2,n
2n,1n ,,,1,,,2,:1,xS,1,x,x,……,x,nxn
nnxnx1,,,xS,1时,,, n21,x,,1,x
,1,,nn,1时,,1,2,3,……,, xSnn2
(3)倒序相加法:把数列的各项顺序倒写,再与原来顺序的数列相加。 Saaaa,,,,,,……nnn12,1相加 ,,,,,,Saaaa……nnn,121,
,,,,,,2S,a,a,a,a,……,a,a……n1n2n,11n
,练习,
2,,,,,,x111,,,,,,已知fx(),,则f(1),f(2),f,f(3),f,f(4),f,,,,,,,22341,x,,,,,,
2,,1,,,,22xxx,,11,,fxf,,(由(),,,,,,1 ,,2222x1,1,1,xxx,,,,1,,1,,,x,,
,,,,,,,,,,,,111,,,,,, ?原式,f(1),f(2),f,f(3),f,f(4),f,,,,,,,,,,,,234,,,,,,,,,,,,
11 ,,,,,1113)22