Question

5£®ÉèͬʱÂú×ãÌõ¼þ£º¢Ù$\frac{{b}_{n}+{b}_{n+2}}{2}$¡Ýb_n+1£»¢Úb_n¡ÜM£¨n¡ÊN^*£¬MÊÇÓëÎ޹صij£Êý£©µÄÎÞÇîÊýÁÐ{b_n}½Ð¡°ºêʵ¡±ÊýÁУ®ÒÑÖªÊýÁÐ{a_n}µÄÇ°ÏîºÍS_nÂú×㣺S_n=$\frac{a}{a-1}$£¨a_n-1£©£¨aΪ³£Êý£¬ÇÒa¡Ù0£¬a¡Ù1£©£®
£¨¢ñ£©Çó{a_n}µÄͨÏʽ£»
£¨¢ò£©Éèb_n=$\frac{2{S}_{n}}{{a}_{n}}$+1£¬ÈôÊýÁÐ{b_n}ΪµÈ±ÈÊýÁУ¬ÇóaµÄÖµ£¬²¢Ö¤Ã÷´Ëʱ{$\frac{1}{{b}_{n}}$}Ϊ¡°ºêʵ¡±ÊýÁУ®

Answer 1

·ÖÎö £¨1£©µ±n¡Ý2ʱ£¬${a_n}={S_n}-{S_{n-1}}=\frac{a}{a-1}{a_n}-\frac{a}{a-1}{a_{n-1}}$$\frac{a_n}{{{a_{n-1}}}}=a$£¬´Ó¶ø¿ÉµÃ{a_n}ÒÔaΪÊ×ÏaΪ¹«±ÈµÄµÈ±ÈÊýÁУ¬ÓÉ´Ë¿ÉÇó{a_n}µÄͨÏʽ£»
£¨2£©È·¶¨ÊýÁÐ{b_n}µÄͨÏÀûÓÃ{b_n}ΪµÈ±ÈÊýÁУ¬¿ÉÇóaµÄÖµ£»ÑéÖ¤¡°ºêʵ¡±ÊýÁеÄÁ½¸öÌõ¼þ£¬¼´¿ÉÖ¤µÃ£®

½â´ð ½â£º£¨1£©ÒòΪ${S_1}=\frac{a}{a-1}£¨{a_1}-1£©$£¬ËùÒÔa₁=a
µ±n¡Ý2ʱ£¬${a_n}={S_n}-{S_{n-1}}=\frac{a}{a-1}{a_n}-\frac{a}{a-1}{a_{n-1}}$$\frac{a_n}{{{a_{n-1}}}}=a$£¬
¼´{a_n}ÒÔaΪÊ×ÏaΪ¹«±ÈµÄµÈ±ÈÊýÁУ®
ËùÒÔ${a_n}=a•{a^{n-1}}={a^n}$£»         ¡­£¨4·Ö£©
£¨2£©ÓÉ£¨1£©Öª£¬${b_n}=\frac{{2¡Á\frac{a}{a-1}£¨{a_n}-1£©}}{a_n}+1=\frac{{£¨3a-1£©{a_n}-2a}}{{£¨a-1£©{a_n}}}$£¬
Èô{b_n}ΪµÈ±ÈÊýÁУ¬ÔòÓÐ${b_2}^2={b_1}•{b_3}$£¬¶øb₁=3£¬${b_2}=\frac{3a+2}{a}$£¬${b_3}=\frac{{3{a^2}+2a+2}}{a^2}$
¹Ê${£¨\frac{3a+2}{a}£©^2}=3•\frac{{3{a^2}+2a+2}}{a^2}$£¬½âµÃ$a=\frac{1}{3}$¡­£¨7·Ö£©
ÔÙ½«$a=\frac{1}{3}$´úÈëµÃ£º${b_n}={3^n}$£¬ÆäΪµÈ±ÈÊýÁУ¬
ËùÒÔ$a=\frac{1}{3}$³ÉÁ¢¡­£¨8·Ö£©
ÓÉÓÚ¢Ù$\frac{{\frac{1}{b_n}+\frac{1}{{{b_{n+2}}}}}}{2}=\frac{{\frac{1}{3^n}+\frac{1}{{{3^{n+2}}}}}}{2}£¾\frac{{2\sqrt{\frac{1}{3^n}•\frac{1}{{{3^{n+2}}}}}}}{2}=\frac{1}{{{3^{n+1}}}}=\frac{1}{{{b_{n+1}}}}$¡­£¨10·Ö£©
£¨»ò×ö²î¸ü¼òµ¥£ºÒòΪ$\frac{{\frac{1}{b_n}+\frac{1}{{{b_{n+2}}}}}}{2}-\frac{1}{{{b_{n+1}}}}=\frac{5}{{{3^{n+2}}}}-\frac{1}{{{3^{n+1}}}}=\frac{2}{{{3^{n+2}}}}£¾0$£¬
ËùÒÔ$\frac{{\frac{1}{b_n}+\frac{1}{{{b_{n+2}}}}}}{2}¡Ý\frac{1}{{{b_{n+1}}}}$Ò²³ÉÁ¢£©
¢Ú$\frac{1}{b_n}=\frac{1}{3^n}¡Ü\frac{1}{3}$£¬¹Ê´æÔÚ$M¡Ý\frac{1}{3}$£»
ËùÒÔ·ûºÏ¢Ù¢Ú£¬¹Ê$\left\{{\frac{1}{b_n}}\right\}$Ϊ¡°¼ÎÎÄ¡±ÊýÁС­£¨12·Ö£©

µãÆÀ ±¾Ì⿼²éµÝÍÆÊýÁеÄÓ¦Óã¬ÒÔ¼°µÈ±ÈÊýÁеĶ¨ÒåÓëͨÏ¿¼²éж¨Ò壬½âÌâµÄ¹Ø¼üÊÇÀí½âж¨Ò壬ÕýÈ·ÔËÓÃж¨Ò壬ÔËËãÁ¿´ó£¬ÓÐÒ»¶¨µÄÄѶÈ