Question

ÉèÊýÁÐ{x_n}µÄËùÓÐÏÊDz»µÈÓÚ1µÄÕýÊý,Ç°nÏîºÍΪS_n,ÒÑÖªµãP_n(x_n,S_n)ÔÚÖ±Ïßy=kx+bÉÏ(ÆäÖÐ,³£Êýk¡Ù0,ÇÒk¡Ù1),ÓÖy_n=log_0.5x_n.

(1)ÇóÖ¤:ÊýÁÐ{x_n}ÊǵȱÈÊýÁÐ;

(2)Èç¹ûy_n=18-3n,ÇóʵÊýk,bµÄÖµ;

(3)Èç¹û´æÔÚt,s¡ÊN^*,s¡Ùt£¬Ê¹µÃµã£¨t,y_s£©ºÍ£¨s,y_t£©¶¼ÔÚÖ±Ïßy=2x+1ÉÏ,ÊÔÅжÏ,ÊÇ·ñ´æÔÚ×ÔÈ»ÊýM,µ±n£¾Mʱ,x_n£¾1ºã³ÉÁ¢?Èô´æÔÚ,Çó³öMµÄ×îСֵ,Èô²»´æÔÚ,Çë˵Ã÷ÀíÓÉ.

Answer 1

½â:(1)Ö¤Ã÷£º¡ßµãP_n,P_n+1¶¼ÔÚÖ±Ïßy=kx+bÉÏ,¡à=k,µÃ(k-1)x_n+1=kx_n.

¡ß³£Êýk¡Ù0,ÇÒk¡Ù1,¡à(·ÇÁã³£Êý).¡àÊýÁÐ{x_n}ÊǵȱÈÊýÁÐ.

(2)ÓÉy_n=log_0.5x_n,µÃx_n==8^n-6=8^-58^n-1,¡à=8,µÃk=.

ÓÉP_nÔÚÖ±ÏßÉÏ,µÃS_n=kx_n+b,Áîn=1µÃb=S₁x₁=x₁=.

(3)x_n£¾1ºã³ÉÁ¢µÈ¼ÛÓÚy_n£¼0,

¡ß´æÔÚt,s¡ÊN£¬Ê¹µÃ(t,y_s)ºÍ£¨s,y_t)¶¼ÔÚy=2x+1ÉÏ,¡ày_s=2t+1,¢Ù

y_t=2s+1,¢Ú

¢Ù-¢Ú,µÃy_s-y_t=2(t-s).Ò×Ö¤{y_n}ÊǵȲîÊýÁÐ,ÉèÆ乫²îΪd,ÔòÓÐy_s-y_t=(s-t)d,

¡ßs¡Ùt,¡àd=-2£¼0.¢Ù+¢Ú,µÃy_s+y_t=2(t+s)+2,

ÓÖy_s+y_t=y₁+(s-1)(-2)+y₁+(t-1)(-2)=2y₁-2(s+t)+4,

ÓÉ2y₁-2(s+t)+4=2(t+s)+2,µÃy₁=2(t+s)-1£¾0,

¼´ÊýÁÐ{y_n}ÊÇÊ×ÏîΪÕý,¹«²îΪ¸ºµÄµÈ²îÊýÁÐ,

¡àÒ»¶¨´æÔÚÒ»¸ö×îС×ÔÈ»ÊýM,ʹ

¼´½âµÃt+s£¼M¡Üt+s+.¡ßM¡ÊN,¡àM=t+s,

¼´´æÔÚ×ÔÈ»ÊýM,Æä×îСֵΪt+s,ʹµÃµ±n£¾Mʱ,x_n£¾1ºã³ÉÁ¢.