Question

6£®ÒÑÖªº¯Êý$f£¨x£©=\frac{x}{1+{2{x^2}}}$£¬¶¨ÒåÕýÊýÊýÁÐ{a_n}£¬${a_1}=\frac{1}{2}$£¬a_n+1²=2a_nf£¨a_n£©£¨n¡ÊN^*£©£®
£¨1£©Ö¤Ã÷ÊýÁÐ$\{\frac{1}{a_n^2}-2\}$ÊǵȱÈÊýÁУ»
£¨2£©Áî${b_n}=\frac{1}{a_n^2}-2$£¬S_nΪÊýÁÐ{b_n}µÄÇ°nÏîºÍ£¬Çóʹ${S_n}£¾\frac{31}{8}$³ÉÁ¢µÄ×îСnµÄÖµ£®

Answer 1

·ÖÎö £¨1£©Í¨¹ý¶Ôa_n+1²=2a_nf£¨a_n£©=$\frac{2{{a}_{n}}^{2}}{1+2{{a}_{n}}^{2}}$Á½±ßͬʱȡµ¹Êý¿ÉÖª$\frac{1}{{{a}_{n+1}}^{2}}$=$\frac{1}{2}$•$\frac{1}{{{a}_{n}}^{2}}$+1£¬ÕûÀí¿ÉÖª$\frac{1}{{{a}_{n+1}}^{2}}$-2=$\frac{1}{2}$•£¨$\frac{1}{{{a}_{n}}^{2}}$-2£©£¬½ø¶ø¿ÉÖªÊýÁÐ$\{\frac{1}{a_n^2}-2\}$ÊÇÒÔ2ΪÊ×Ïî¡¢$\frac{1}{2}$Ϊ¹«±ÈµÄµÈ±ÈÊýÁУ»
£¨2£©Í¨¹ý£¨1£©¿ÉÖªb_n=$\frac{1}{{2}^{n-2}}$£¬ÀûÓõȱÈÊýÁеÄÇóºÍ¹«Ê½¿ÉÖªS_n=4£¨1-$\frac{1}{{2}^{n}}$£©£¬´Ó¶ø${S_n}£¾\frac{31}{8}$µÈ¼ÛÓÚ4£¨1-$\frac{1}{{2}^{n}}$£©£¾$\frac{31}{8}$£¬½ø¶ø¼ÆËã¿ÉµÃ½áÂÛ£®

½â´ð £¨1£©Ö¤Ã÷£ºÒÀÌâÒ⣬a_n+1²=2a_nf£¨a_n£©=2a_n•$\frac{{a}_{n}}{1+2{{a}_{n}}^{2}}$=$\frac{2{{a}_{n}}^{2}}{1+2{{a}_{n}}^{2}}$£¬
¡à$\frac{1}{{{a}_{n+1}}^{2}}$=$\frac{1+2{{a}_{n}}^{2}}{2{{a}_{n}}^{2}}$=$\frac{1}{2}$•$\frac{1}{{{a}_{n}}^{2}}$+1£¬
ÕûÀíµÃ£º$\frac{1}{{{a}_{n+1}}^{2}}$-2=$\frac{1}{2}$•£¨$\frac{1}{{{a}_{n}}^{2}}$-2£©£¬
ÓÖ¡ß$\frac{1}{{{a}_{1}}^{2}}$-2=$\frac{1}{£¨\frac{1}{2}£©^{2}}$-2=2£¬
¡àÊýÁÐ$\{\frac{1}{a_n^2}-2\}$ÊÇÒÔ2ΪÊ×Ïî¡¢$\frac{1}{2}$Ϊ¹«±ÈµÄµÈ±ÈÊýÁУ»
£¨2£©½â£ºÓÉ£¨1£©¿ÉÖª${b_n}=\frac{1}{a_n^2}-2$=2•$\frac{1}{{2}^{n-1}}$=$\frac{1}{{2}^{n-2}}$£¬
¡àS_n=$\frac{\frac{1}{{2}^{-1}}•£¨1-\frac{1}{{2}^{n}}£©}{1-\frac{1}{2}}$
=4£¨1-$\frac{1}{{2}^{n}}$£©£¬
¡ß${S_n}£¾\frac{31}{8}$£¬¼´4£¨1-$\frac{1}{{2}^{n}}$£©£¾$\frac{31}{8}$£¬
ÕûÀíµÃ£º$\frac{1}{{2}^{n}}$£¼$\frac{1}{32}$=$\frac{1}{{2}^{5}}$£¬
¡àn£¾5£¬
¡àÂú×ãÌõ¼þµÄn×îСֵΪ6£®

µãÆÀ ±¾Ì⿼²éÊýÁеÄͨÏî¼°Ç°nÏîºÍ£¬¶Ô±í´ïʽµÄÁé»î±äÐÎÊǽâ¾ö±¾ÌâµÄ¹Ø¼ü£¬×¢Òâ½âÌâ·½·¨µÄ»ýÀÛ£¬ÊôÓÚÖеµÌ⣮