ãã(ä¸)1.çå·®æ°å{an}:
é项å
¬å¼an=a1+(n-1)d é¦é¡¹a1,å
¬å·®d, an第n项æ°
an=ak+(n-k)d ak为第k项æ°
è¥a,A,bææçå·®æ°å å A=(a+b)/2
2.çå·®æ°åån项å:
设çå·®æ°å{an}çån项å为Sn
å³ Sn=a1+a2+...+an;
é£ä¹ Sn=na1+n(n-1)d/2
=dn^2(å³nç2次æ¹) /2+(a1-d/2)n
è¿æ以ä¸çæ±åæ¹æ³: 1,ä¸å®å
¨å½çº³æ³ 2 ç´¯å æ³ 3 ååºç¸å æ³
(äº)1.çæ¯æ°å{an}:
é项å
¬å¼ an=a1*q^(n-1)(å³qçn-1次æ¹) a1为é¦é¡¹,an为第n项
an=a1*q^(n-1,am=a1*q^(m-1))
åan/am=q^(n-m)
(1)an=am*q^(n-m)
(2)a,G,b è¥ææçæ¯ä¸é¡¹,åG^2=ab (a,b,Gä¸çäº0)
(3)è¥m+n=p+q å am*an=ap+aq
2.çæ¯æ°åån项å{an}
设 a1,a2,a3...anææçæ¯æ°å
ån项åSn=a1+a2+a3...an
Sn=a1+a1*q+a1*q^2+....a1*q^(n-2)+a1*q^(n-1)(è¿ä¸ªå
¬å¼è½ç¶æ¯æåºæ¬å
¬å¼,ä½ä¸é¨åé¢ç®ä¸æ±ån项åæ¯å¾é¾ç¨ä¸é¢é£ä¸ªå
¬å¼æ¨åç,è¿æ¶å¯è½è¦ç´æ¥ä»åºæ¬å
¬å¼æ¨åè¿å»,æ以å¸æä½ è¿ä¸ªå
¬å¼ä¹è¦ç解)
Sn=a1(1-q^n)/(1-q)=(a1-an*q)/(1-q);
注: qä¸çäº1;
Sn=na1 注:q=1
æ±åä¸è¬æ以ä¸4个æ¹æ³: 1,ä¸å®å
¨å½çº³æ³ 2 ç´¯ä¹æ³ 3 éä½æ±åæ³