Question

20£®¶ÔÓÚÊýÁÐ{a_n}£¬³Æ$P£¨{a_k}£©=\frac{1}{k-1}£¨|{{a_1}-{a_2}}|+|{{a_2}-{a_3}}|+¡­+|{{a_{k-1}}-{a_k}}|£©$£¨ÆäÖÐk¡Ý2£¬k¡ÊN£©ÎªÊýÁÐ{a_n}µÄÇ°kÏî¡°²¨¶¯¾ùÖµ¡±£®Èô¶ÔÈÎÒâµÄk¡Ý2£¬k¡ÊN£¬¶¼ÓÐP£¨a_k+1£©£¼P£¨a_k£©£¬Ôò³ÆÊýÁÐ{a_n}Ϊ¡°Ç÷ÎÈÊýÁС±£®
£¨1£©ÈôÊýÁÐ1£¬x£¬2Ϊ¡°Ç÷ÎÈÊýÁС±£¬ÇóxµÄÈ¡Öµ·¶Î§£»
£¨2£©Èô¸÷Ïî¾ùΪÕýÊýµÄµÈ±ÈÊýÁÐ{b_n}µÄ¹«±Èq¡Ê£¨0£¬1£©£¬ÇóÖ¤£º{b_n}ÊÇ¡°Ç÷ÎÈÊýÁС±£»
£¨3£©ÒÑÖªÊýÁÐ{a_n}µÄÊ×ÏîΪ1£¬¸÷Ïî¾ùΪÕûÊý£¬Ç°kÏîµÄºÍΪS_k£®ÇÒ¶ÔÈÎÒâk¡Ý2£¬k¡ÊN£¬¶¼ÓÐ3P£¨S_k£©=2P£¨a_k£©£¬ÊÔ¼ÆË㣺$C_n^2P£¨{a_2}£©+2C_n^3P£¨{a_3}£©+¡­+£¨n-1£©C_n^nP£¨{a_n}£©$£¨n¡Ý2£¬n¡ÊN£©£®

Answer 1

·ÖÎö £¨1£©ÓÉж¨Òå¿ÉµÃ|1-x|£¾$\frac{|1-x|+|x-2|}{2}$£¬½â²»µÈʽ¿ÉµÃxµÄ·¶Î§£»
£¨2£©ÔËÓõȱÈÊýÁеÄͨÏʽºÍÇóºÍ¹«Ê½£¬½áºÏж¨Ò壬ÔËÓò»µÈʽµÄÐÔÖʼ´¿ÉµÃÖ¤£»
£¨3£©ÓÉÈÎÒâk¡Ý2£¬k¡ÊN£¬¶¼ÓÐ3P£¨S_k£©=2P£¨a_k£©£¬¿ÉµÃ£¨k-1£©P£¨S_k£©-£¨k-2£©P£¨S_k-1£©=|a_k|£¬ÓɵȱÈÊýÁеÄͨÏʽ£¬¿ÉµÃ${a_k}={£¨-2£©^{k-1}}$£¬½áºÏж¨ÒåºÍ¶þÏîʽ¶¨Àí£¬»¯¼òÕûÀí¼´¿ÉµÃµ½ËùÇóÖµ£®

½â´ð ½â£º£¨1£©ÓÉÌâÒâ¿ÉµÃ$|{1-x}|£¾\frac{{|{1-x}|+|{x-2}|}}{2}$£¬
¼´|1-x|£¾|x-2|£¬Á½±ßƽ·½¿ÉµÃx²-2x+1£¾x²-4x+4£¬
½âµÃ$x£¾\frac{3}{2}$£»
£¨2£©Ö¤Ã÷£ºÓÉÒÑÖª£¬Éè${b_n}={b_1}{q^{n-1}}£¨{b_1}£¾0£©$£¬
Òòb₁£¾0ÇÒ0£¼q£¼1£¬
¹Ê¶ÔÈÎÒâµÄk¡Ý2£¬k¡ÊN*£¬¶¼ÓÐb_k-1£¾b_k£¬
¡à$P£¨{b_k}£©=\frac{1}{k-1}£¨|{{b_1}-{b_2}}|+|{{b_2}-{b_3}}|+¡­+|{{b_{k-1}}-{b_k}}|£©$
=$\frac{1}{k-1}£¨{b_1}-{b_2}+{b_2}-{b_3}+¡­+{b_{k-1}}-{b_k}£©=\frac{{{b_1}£¨1-q£©}}{k-1}£¨1+q+{q^2}+¡­+{q^{k-2}}£©$£¬$P£¨{b_{k+1}}£©=\frac{{{b_1}£¨1-q£©}}{k}£¨1+q+{q^2}+¡­+{q^{k-1}}£©$£¬
Òò0£¼q£¼1¡àqⁱ£¾q^k-1£¨i£¼k-1£©£¬
¡à1£¾q^k-1£¬q£¾q^k-1£¬q²£¾q^k-1£¬¡­£¬q^k-2£¾q^k-1£¬
¡à1+q+q²+¡­+q^k-2£¾£¨k-1£©q^k-1£¬
¡àk£¨1+q+q²+¡­+q^k-2£©£¾£¨k-1£©£¨1+q+q²+¡­+q^k-2+q^k-1£©£¬
¡à$\frac{{£¨1+q+{q^2}+¡­+{q^{k-2}}£©}}{k-1}£¾\frac{{£¨1+q+{q^2}+¡­+{q^{k-2}}+{q^{k-1}}£©}}{k}$£¬
¡à$\frac{{{b_1}£¨1-q£©£¨1+q+{q^2}+¡­+{q^{k-2}}£©}}{k-1}£¾\frac{{{b_1}£¨1-q£©£¨1+q+{q^2}+¡­+{q^{k-2}}+{q^{k-1}}£©}}{k}$
¼´¶ÔÈÎÒâµÄk¡Ý2£¬k¡ÊN*£¬¶¼ÓÐP£¨b_k£©£¾P£¨b_k+1£©£¬¹Ê{b_n}ÊÇ¡°Ç÷ÎÈÊýÁС±£»
£¨3£©µ±k¡Ý2ʱ£¬$P£¨{S_k}£©=\frac{1}{k-1}£¨|{{S_1}-{S_2}}|+|{{S_2}-{S_3}}|+¡­+|{{S_{k-1}}-{S_k}}|£©=\frac{1}{k-1}£¨|{a_2}|+|{a_3}|+¡­+|{a_k}|£©$
µ±k¡Ý3ʱ£¬$P£¨{S_{k-1}}£©=\frac{1}{k-2}£¨|{a_2}|+|{a_3}|+¡­+|{{a_{k-1}}}|£©$£¬
¡à£¨k-1£©P£¨S_k£©-£¨k-2£©P£¨S_k-1£©=|a_k|
ͬÀí£¬£¨k-1£©P£¨a_k£©-£¨k-2£©P£¨a_k-1£©=|a_k-1-a_k|£¬
Òò3P£¨S_k£©=2P£¨a_k£©£¬
¡à3£¨k-1£©P£¨S_k£©=2£¨k-1£©P£¨a_k£©3£¨k-2£©P£¨S_k-1£©=2£¨k-2£©P£¨a_k-1£©£¬
¼´3|a_k|=2|a_k-1-a_k|£¬
ËùÒÔ3a_k=2£¨a_k-1-a_k£©»ò  3a_k=-2£¨a_k-1-a_k£©
ËùÒÔ  a_k=-2a_k-1»ò  ${a_k}=\frac{2}{5}{a_{k-1}}$
ÒòΪa₁=1£¬ÇÒa_k¡ÊZ£¬ËùÒÔa_k=-2a_k-1£¬´Ó¶ø${a_k}={£¨-2£©^{k-1}}$£¬
ËùÒÔp£¨a_k£©=$\frac{1}{k-1}$£¨|1-£¨-2£©|+|-2-£¨-2£©²|+¡­+|£¨-2£©^k-2-£¨-2£©^k-1£©=$\frac{3£¨{2}^{k-1}-1£©}{k-1}$£¬$C_n^2P£¨{a_2}£©+2C_n^3P£¨{a_3}£©+3C_n^4P£¨{a_4}£©+¡­+£¨n-1£©C_n^nP£¨{a_n}£©$
=$3[£¨C_n^2•2+C_n^3•{2^2}+C_n^4•{2^3}+¡­+C_n^n•{2^{n-1}}£©-£¨C_n^2+C_n^3+C_n^4+¡­+C_n^n£©]$
=$3[\frac{1}{2}£¨{3^n}-2n-1£©-£¨{2^n}-n-1£©]=\frac{3}{2}£¨{3^n}-{2^{n+1}}+1£©$£®

µãÆÀ ±¾Ì⿼²éж¨ÒåµÄÀí½âºÍÔËÓ㬿¼²éµÈ±ÈÊýÁеÄͨÏʽºÍÇóºÍ¹«Ê½µÄÔËÓ㬿¼²é¶þÏîʽ¶¨ÀíµÄÔËÓ㬿¼²éÔËËãÄÜÁ¦£¬ÊôÓÚÄÑÌ⣮