文档介绍：高中数学数列知识点总结(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a10,da≤0⎩n+1⎧a≤0当a10,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: