ÌâÄ¿ÄÚÈÝ

20£®ÒÑÖªµÈ²îÊýÁÐ{an}µÄÊ×ÏîΪc£¬¹«²îΪd£¬µÈ±ÈÊýÁÐ{bn}µÄÊ×ÏîΪd£¬¹«±ÈΪc£¬ÆäÖÐc£¬d¡ÊZ£¬ÇÒa1£¼b1£¼a2£¼
b2£¼a3£®
£¨1£©ÇóÖ¤£º0£¼c£¼d£¬²¢ÓÉb2£¼a3ÍÆµ¼cµÄÖµ£»
£¨2£©ÈôÊýÁÐ{an}¹²ÓÐ3nÏǰnÏîµÄºÍΪA£¬ÆäºóµÄnÏîµÄºÍΪB£¬ÔÙÆäºóµÄnÏîµÄºÍΪC£¬Çó$\frac{{B}^{2}-AC}{£¨A-C£©^{2}}$µÄ±ÈÖµ£®
£¨3£©ÈôÊýÁÐ{bn}µÄǰnÏǰ2nÏǰ3nÏîµÄºÍ·Ö±ðΪD£¬G£¬H£¬ÊÔÓú¬×ÖĸD£¬GµÄʽ×ÓÀ´±íʾH£¨¼´H=f£¨D£¬G£©£¬ÇÒ²»º¬×Öĸd£©

·ÖÎö £¨1£©¸ù¾ÝµÈ²î¡¢µÈ±ÈÊýÁеÄͨÏʽ¿ÉÒÔÍÆÖª0£¼c£¼d£¬½áºÏÒÑÖªÌõ¼þa1£¼b1£¼a2£¼b2£¼a3Áгö²»µÈʽ×飺$\left\{\begin{array}{l}{0£¼c£¼d}\\{d£¼cd}\\{cd£¼c+2d£¼3d}\end{array}\right.$£¬Í¨¹ý½â¸Ã²»µÈʽ×éÍÆµ¼cµÄÖµ£»
£¨2£©¸ù¾ÝµÈ²îÊýÁеÄͨÏʽºÍÐÔÖÊÍÆÖªA=Sn£¬B=S2n-Sn£¬C=S3n-S2n£¬Ò×µÃB¡¢A+C=2B£¬½áºÏ´úÊýʽµÄ±äÐÎÀ´Çó$\frac{{B}^{2}-AC}{£¨A-C£©^{2}}$µÄÖµ£»
£¨3£©¸ù¾ÝµÈ±ÈÊýÁеÄǰnÏîºÍ¹«Ê½·Ö±ð±íʾ³öD¡¢G¡¢H£¬È»ºóÕÒµ½ËüÃǵÄÊýÁ¿¹ØÏµ£®

½â´ð ½â£º£¨1£©ÒÑÖªa1=c£¬a2=c+d£¬a3=c+2d£¬b1=d£¬b2=dc£¬
ÓÉb1£¼a2¿ÉÖªc£¾0£¬Òò´Ë0£¼c£¼d£¬
ÓÉa1£¼b1£¼a2£¼b2£¼a3¿ÉµÃ£ºc£¼d£¼c+d£¼cd£¼c+2d£¬ÇÒc£¬d¡ÊZ£¬
Òò´Ë¿ÉµÃ²»µÈʽ×飺$\left\{\begin{array}{l}{0£¼c£¼d}\\{d£¼cd}\\{cd£¼c+2d£¼3d}\end{array}\right.$⇒$\left\{\begin{array}{l}{0£¼c}\\{1£¼c}\\{c£¼3}\end{array}\right.$⇒1£¼c£¼3£®
ÓÖÒòΪc¡ÊZ£¬
Òò´Ëc=2£»
£¨2£©ÊýÁÐ{an}µÄͨÏîΪÊýÁÐan=2+£¨n-1£©d£¬Sn=$\frac{d}{2}$n2+£¨2-$\frac{d}{2}$£©n£¬A=Sn£¬B=S2n-Sn£¬C=S3n-S2n£¬
B=$\frac{d}{2}$£¨4n2-n2£©+£¨2-$\frac{d}{2}$£©£¨2n-n£©=$\frac{d}{2}$•3n2+£¨2-$\frac{d}{2}$£©n£¬
¿ÉµÃA+C=$\frac{d}{2}$n2+£¨2-$\frac{d}{2}$£©n+$\frac{d}{2}$£¨9n2-4n2£©+£¨2-$\frac{d}{2}$£©£¨3n-2n£©=3d•n2+£¨2-$\frac{d}{2}$£©•2n£¬
¿ÉµÃA+C=2B£¬
Òò´Ë$\frac{{B}^{2}-AC}{£¨A-C£©^{2}}$=$\frac{£¨A+C£©^{2}-4AC}{4£¨A-C£©}$=$\frac{1}{4}$£»
£¨3£©ÊýÁÐ{bn}µÄͨÏîΪbn=d•2n-1£®
Òò´ËD=$\frac{d£¨{2}^{n}-1£©}{2-1}$=d£¨2n-1£©£¬G=d£¨22n-1£©£¬H=d£¨23n-1£©£®
ËùÒÔ$\left\{\begin{array}{l}{G=£¨{2}^{n}+1£©•D}\\{H=£¨{2}^{3n}+{2}^{n}+1£©•D}\end{array}\right.$£¬
Òò´ËH=D•£¨$\frac{G}{D}$-1£©2+G=$\frac{{G}^{2}}{D}$+2D-G£®

µãÆÀ ±¾Ì⿼²éµÈ±È¡¢µÈ²îÊýÁеÄͨÏʽ¼°Ó¦Óã¬ÊýÁеÄÇóºÍ£¬¿¼²é¼ÆËãÄÜÁ¦£¬ÊôÓÚÄѶȽϴóµÄÌâÄ¿£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø