Question

2£®ÒÑÖªÊýÁÐ{a_n}ÖУ¬a₁=t£¨t¡Ù0ÇÒt¡Ù1£©£¬a₂=t²£¬ÇÒµ±x=tʱ£¬º¯Êýf£¨x£©=$\frac{1}{2}$£¨a_n-a_n-1£©x²-£¨a_n+1-a_n£©x£¨n¡Ý2£¬n¡ÊN^*£©È¡µÃ¼«Öµ£®
£¨1£©ÇóÖ¤£ºÊýÁÐ{a_n+1-a_n}ÊǵȱÈÊýÁУ»
£¨2£©ÇóÊýÁÐ{a_n]µÄͨÏʽ£»
£¨3£©µ±t=-$\sqrt{\frac{7}{10}}$ʱ£¬Èôb_n=a_nln|a_n|£¬ÊýÁÐ{b_n}ÖÐÊÇ·ñ´æÔÚ×î´óÏÈç¹û´æÔÚ£¬ËµÃ÷Êǵڼ¸ÏÈç¹û²»´æÔÚ£¬ËµÃ÷ÀíÓÉ£®

Answer 1

·ÖÎö £¨1£©¸ù¾Ýµ±x=tʱ£¬f£¨x£©=$\frac{1}{2}$£¨a_n-a_n-1£©x²-£¨a_n+1-a_n£©x£¨n¡Ý2£©È¡µÃ¼«Öµ£¬Çóµ¼£¬µÃµ½f'£¨t£©=0£¬¼´a_n-a_n-1£©t=a_n+1-a_n£¨n¡Ý2£©ÕûÀí¿ÉÖ¤£»
£¨2£©Í¨¹ý£¨1£©¡¢ÀûÓÃÀÛ¼Ó·¨¼´¿ÉÇóµÃÊýÁÐ{a_n}µÄͨÏʽ£»
£¨3£©¸ù¾Ý£¨2£©È¥¾ø¶ÔÖµ·ûºÅ¿ÉÇóÊýÁÐ{b_n}µÄͨÏʽ£¬¶Ôn·ÖÆæżÌÖÂÛ¼´µÃ½áÂÛ£®

½â´ð £¨1£©Ö¤Ã÷£ºÁîf¡ä£¨t£©=£¨a_n-a_n-1£©t-£¨a_n+1-a_n£©=0£¬
µÃ£º£¨a_n-a_n-1£©t=a_n+1-a_n£¨n¡Ý2£©£¬
ÓÖa₂-a₁=t£¨t-1£©£¬t¡Ù0ÇÒt¡Ù1£¬
¡àa₂-a₁¡Ù0£¬
¡à$\frac{{a}_{n+1}-{a}_{n}}{{a}_{n}-{a}_{n-1}}$=t£¬
¡àÊýÁÐ{a_n+1-a_n}ÊÇÊ×ÏîΪt²-t¡¢¹«±ÈΪtµÄµÈ±ÈÊýÁУ»
£¨2£©½â£ºÓÉ£¨1£©Öªa_n+1-a_n=tⁿ⁺¹-tⁿ£¬
¡àa_n-a_n-1=tⁿ-t^n-1£¬
¡àa_n-1-a_n-2=t^n-1-t^n-2£¬
¡­
a₂-a₁=t²-t£¬
ÉÏÃæn-1¸öµÈʽÏà¼Ó²¢ÕûÀíµÃ£ºa_n=tⁿ£¨t¡Ù0ÇÒt¡Ù1£©£»
£¨3£©½áÂÛ£ºÊýÁÐ{b_n}ÖеÄ×î´óÏîΪµÚ5Ï
ÀíÓÉÈçÏ£º
ÓÉ£¨2£©Öªb_n=a_nln|a_n|=tⁿ•ln|tⁿ|=ntⁿ•ln|t|£®
¡ßt=-$\sqrt{\frac{7}{10}}$£¬¡à-1£¼t£¼0£¬
¡ßµ±nΪżÊýʱb_n£¾0£¬µ±nΪÆæÊýʱb_n£¼0£¬
¡à×î´óÏî±ØÐëΪÆæÊýÏ
Éè×î´óÏîΪ£ºb_2k+1£¬Ôò$\left\{\begin{array}{l}{{b}_{2k+1}¡Ý{b}_{2k-1}}\\{{b}_{2k+1}¡Ý{b}_{2k+3}}\end{array}\right.$£¬
¡à$\left\{\begin{array}{l}{£¨2k+1£©•{t}^{2k+1}•ln|t|¡Ý£¨2k-1£©•{2}^{2k-1}•ln|t|}\\{£¨2k+1£©•{t}^{2k+1}•ln|t|¡Ý£¨2k+3£©•{t}^{2k+3}•ln|t|}\end{array}\right.$£¬
ÕûÀíµÃ£º$\left\{\begin{array}{l}{£¨2k+1£©•{t}^{2}¡Ý2k-1}\\{2k+1¡Ý£¨2k+3£©•{t}^{2}}\end{array}\right.$£¬
½«t=-$\sqrt{\frac{7}{10}}$´úÈëÉÏʽ£¬½âµÃ£º$\frac{11}{6}$¡Ük¡Ü$\frac{17}{6}$£¬
¡àk=2£¬¼´ÊýÁÐ{b_n}ÖеÄ×î´óÏîΪµÚ5Ï

µãÆÀ ±¾Ì⿼²éµÈ±ÈÊýÁеĶ¨ÒåºÍͨÏʽ£¬ÀÛ¼Ó·¨ÇóÊýÁÐͨÏʽ£¬ÌåÏÖÁË·ÖÀàÌÖÂÛµÄ˼Ï룮ÆäÖÐÎÊÌ⣨3£©ÊÇÒ»¸ö¿ª·ÅÐÔÎÊÌ⣬¿¼²éÁËͬѧÃǹ۲졢ÍÆÀíÒÔ¼°´´ÔìÐԵطÖÎöÎÊÌâ¡¢½â¾öÎÊÌâµÄÄÜÁ¦£®×¢Òâ½âÌâ·½·¨µÄ»ýÀÛ£¬ÊôÓÚÄÑÌ⣮