Question

17£®ÒÑÖªµÈ²îÊýÁÐ{a_n}µÄÇ°nÏîºÍΪS_n£¬ÇÒa₂=0£¬S₅=10£¬ÊýÁÐ{b_n}Âú×ãb₁=$\frac{1}{2}$£¬$\frac{{b}_{n+1}}{n+1}$=$\frac{1}{2}$•$\frac{{b}_{n}}{n}$£®
£¨¢ñ£©ÇóÊýÁÐ{a_n}£¬{b_n}µÄͨÏʽ£»
£¨¢ò£©¼ÇT_nΪÊýÁÐ{b_n}µÄÇ°nÏîºÍ£¬f£¨n£©=$\frac{£¨{S}_{n}+2£©£¨2-{T}_{n}£©}{n+2}$£¬ÊÔÎÊf£¨n£©ÊÇ·ñ´æÔÚ×î´óÖµ£¬Èô´æÔÚ£¬Çó³ö×î´óÖµ£»Èô²»´æÔÚ£¬Çë˵Ã÷ÀíÓÉ£®

Answer 1

·ÖÎö £¨¢ñ£©ÔËÓõȲîÊýÁеÄÇóºÍ¹«Ê½ºÍͨÏ¿ÉµÃÊýÁÐ{a_n}µÄͨÏÔËÓõȱÈÊýÁеĶ¨Ò壬¿ÉµÃÊýÁÐ{b_n}µÄͨÏʽ£»
£¨¢ò£©ÔËÓôíλÏà¼õ·¨£¬½áºÏµÈ±ÈÊýÁеÄÇóºÍ¹«Ê½£¬¿ÉµÃT_n£¬ÇóµÃf£¨n£©µÄ±í´ïʽ£¬×÷²îf£¨n+1£©-f£¨n£©£¬»¯¼òÕûÀí£¬¿ÉµÃf£¨n£©µÄµ¥µ÷ÐÔ£¬¼´¿ÉµÃµ½×î´óÖµ£®

½â´ð ½â£º£¨¢ñ£©ÓÉS₅=10£¬µÃ5a₃=10£¬¼´a₃=2£¬
ÓÖa₂=0£¬ËùÒÔ¹«²îd=2£¬
Ôòa_n=2n-4£®
ÓÉ$\frac{{b}_{n+1}}{n+1}$=$\frac{1}{2}$•$\frac{{b}_{n}}{n}$Öª£¬
{$\frac{{b}_{n}}{n}$}ÊǵȱÈÊýÁУ¬
Ôò$\frac{{b}_{n}}{n}$=$\frac{{b}_{1}}{1}$•£¨$\frac{1}{2}$£©^n-1£¬
¼´ÓÐb_n=$\frac{n}{{2}^{n}}$£»
£¨¢ò£©ÓÉ£¨¢ñ£©£¬µÃT_n=$\frac{1}{2}$+$\frac{2}{{2}^{2}}$+$\frac{3}{{2}^{3}}$+¡­+$\frac{n}{{2}^{n}}$£¬
Ôò$\frac{1}{2}$T_n=$\frac{1}{{2}^{2}}$+$\frac{2}{{2}^{3}}$+¡­+$\frac{n-1}{{2}^{n}}$+$\frac{n}{{2}^{n+1}}$£¬
Á½Ê½Ïà¼õ£¬»¯¼ò£¬$\frac{1}{2}$T_n=$\frac{1}{2}$+$\frac{1}{{2}^{2}}$+¡­+$\frac{1}{{2}^{n}}$-$\frac{n}{{2}^{n+1}}$
=$\frac{\frac{1}{2}£¨1-\frac{1}{{2}^{n}}£©}{1-\frac{1}{2}}$-$\frac{n}{{2}^{n+1}}$£¬
ÇóµÃT_n=2-$\frac{n+2}{{2}^{n}}$£¬
ÓÉS_n=$\frac{n£¨-2+2n-4£©}{2}$=n²-3n£¬
ËùÒÔf£¨n£©=$\frac{£¨{S}_{n}+2£©£¨2-{T}_{n}£©}{n+2}$=$\frac{{n}^{2}-3n+2}{{2}^{n}}$£¬
Ôòf£¨n+1£©-f£¨n£©=$\frac{£¨n+1£©^{2}-3£¨n+1£©+2}{{2}^{n+1}}$-$\frac{{n}^{2}-3n+2}{{2}^{n}}$£¬
=$\frac{£¨n-1£©£¨4-n£©}{{2}^{n+1}}$£¬
¡àf£¨1£©=f£¨2£©£¼f£¨3£©£¼f£¨4£©=f£¨5£©£¬
µ±n¡Ý5ʱ£¬f£¨n+1£©-f£¨n£©£¼0£¬ÓÖf£¨4£©=f£¨5£©=$\frac{3}{8}$£¬
¡àf£¨n£©´æÔÚ×î´óÖµ£¬×î´óֵΪ$\frac{3}{8}$£®

µãÆÀ ±¾Ì⿼²éµÈ²îÊýÁк͵ȱÈÊýÁеÄͨÏîºÍÇóºÍ¹«Ê½µÄÔËÓã¬Í¬Ê±¿¼²éÊýÁеÄÇóºÍ·½·¨£º´íλÏà¼õ·¨£¬ÒÔ¼°ÊýÁеĵ¥µ÷ÐÔµÄÔËÓãºÇó×îÖµ£¬ÊôÓÚÖеµÌ⣮