高中数学数列笔记

一、等差数列

基础回顾：an+1−an=dan=a1+(n−1)d=am+(n−m)d

Sn=(a1+an)n2=d2n2+(a1−d2)n常

其中an=dn+a₁-d为一次函数Sn=An2+Bn为不含数项的

K-b二次函数

性质：(1)中项性质(充要条件，互推)

an=S2n−12n−1

S2n-1=an(2n-1)

S2n=n(am2n−n+1+am)

↓3=S55

S5=a3×5a3

S4=2(a1+a4)

=2(a₂+a₃)

(2)Snn=d2n+a1−d2数列{Snn}是以d2为公差的等差数列

S3=3S6=9Sn=?⇒.S3=1 06=3/2S1111=?⇒d=32−16−3=16

⇒d=−1∴am+n=an+md=m+mol

(3)an=mam=nan+m=0=m−m=0S1111=1+16(11−3)=73

一般式an+m=nan−mamn−m(n≠m)∴S11=773

等差数列(4) {an}和{bn}前n项和为 Sn和 Tn\frac{S_{2 n - 1}}{T . 2 n - 1}} = \frac{a_{n}}{b_{n}}\frac{S_{2 n - 1}}{T . 2 n - 1}} = \frac{a_{n}}{b_{n}}

(5)项数为2nS偶-S奇=nd可底侧S可底S侧=SnSnanan+1中间两项

项数为⁤2 n - 1 S_{3} = a_{n} S_{1} = ( n = 1 ) a n , S_{合} = n a_{n} . \frac{S_{合}}{S_{1} = \frac{n - 1}⁤

(6)若Sn=mSm=n,则Sm+n=-(n+m)若Sn=Sm,Sm+n=0

一般地，有Sm+n=(n+m)(Sn−Sm)n−m1S2n=2Sn+n2d

b1=a1+dbn=a1+(2n−1)d

⁤\therefore S_{n = d_{n}^{2} + ( a_{1} + d - d ) n}{d n^{2} + a_{1 n}⁤∴S依-S高=dn

S_{A} = a_{n}^{2} + ( a_{1} - d ) n \cdot}S_{A} = a_{n}^{2} + ( a_{1} - d ) n \cdot}S奇=dn+ai-d=a₁+n·1dan

a1+nα=am+1

高中数学数列笔记第1张