·ÖÎö £¨1£©ÔËÓôúÈë·¨£¬¼ÆËã¼´¿ÉµÃµ½ËùÇóa2£¬a3£»
£¨2£©ÓɵȲîÊýÁе͍Ò壬¿ÉµÃ$\frac{{a}_{2}-4k}{{a}_{2}-2}$-$\frac{{a}_{1}-k}{{a}_{1}-1}$=1£¬$\frac{{a}_{3}-9k}{{a}_{3}-3}$-$\frac{{a}_{2}-4k}{{a}_{2}-2}$=1£¬½â·½³Ì¿ÉµÃkµÄÖµ£»
£¨3£©ÔËÓõȲîÊýÁеÄͨÏʽ»¯¼ò¿ÉµÃan=$\frac{2n-{n}^{2}}{n+1}$£¬Çó³öbn=$\frac{1}{{{{£¨{\sqrt{3}}£©}^{n+2}}{a_{n+2}}}}$=$\frac{1}{£¨\sqrt{3}£©^{n+2}•\frac{-n£¨n+2£©}{n+3}}$=$\frac{1}{2}$[$\frac{1}{£¨\sqrt{3}£©^{n+2}£¨n+2£©}$-$\frac{1}{£¨\sqrt{3}£©^{n}n}$]£¬ÔËÓÃÁÑÏîÏàÏûÇóºÍ£¬ÇóµÃÊýÁÐ{bn}µÄǰnÏîºÍΪSn£¬ÔËÓò»µÈʽµÄÐÔÖʼ´¿ÉµÃÖ¤£®
½â´ð ½â£º£¨1£©ÊýÁÐ{an}Âú×ãa1=$\frac{1}{2}$£¬an+1=$\frac{{£¨{n+1}£©£¨{2{a_n}-n}£©}}{{{a_n}+4n}}$£¨n¡ÊN*£©£¬
¿ÉµÃa2=$\frac{2£¨2{a}_{1}-1£©}{{a}_{1}+4}$=0£»a3=$\frac{3£¨2{a}_{2}-2£©}{{a}_{2}+4}$=$\frac{3¡Á£¨0-2£©}{0+8}$=-$\frac{3}{4}$£»
£¨2£©ÊýÁÐ{$\frac{{a}_{n}-k{n}^{2}}{{a}_{n}-n}$}Ϊ¹«²îΪ1µÄµÈ²îÊýÁУ¬
¿ÉµÃ$\frac{{a}_{2}-4k}{{a}_{2}-2}$-$\frac{{a}_{1}-k}{{a}_{1}-1}$=1£¬¼´2k-$\frac{1-2k}{-1}$=1ºã³ÉÁ¢£»
ÓÉ$\frac{{a}_{3}-9k}{{a}_{3}-3}$-$\frac{{a}_{2}-4k}{{a}_{2}-2}$=1£¬¼´$\frac{-\frac{3}{4}-9k}{-\frac{3}{4}-3}$-2k=1£¬
½âµÃk=2£»
£¨3£©Ö¤Ã÷£ºÓÉ£¨2£©¿ÉµÃ$\frac{{a}_{n}-2{n}^{2}}{{a}_{n}-n}$=$\frac{\frac{1}{2}-2}{\frac{1}{2}-1}$+£¨n-1£©=n+2£¬
»¯¼ò¿ÉµÃan=$\frac{2n-{n}^{2}}{n+1}$£¬
bn=$\frac{1}{{{{£¨{\sqrt{3}}£©}^{n+2}}{a_{n+2}}}}$=$\frac{1}{£¨\sqrt{3}£©^{n+2}•\frac{-n£¨n+2£©}{n+3}}$
=-$\frac{n+3}{£¨\sqrt{3}£©^{n+2}n£¨n+2£©}$=$\frac{1}{2}$[$\frac{1}{£¨\sqrt{3}£©^{n+2}£¨n+2£©}$-$\frac{1}{£¨\sqrt{3}£©^{n}n}$]£¬
ÔòǰnÏîºÍΪSn=$\frac{1}{2}$[$\frac{1}{£¨\sqrt{3}£©^{3}•3}$-$\frac{1}{\sqrt{3}•1}$+$\frac{1}{£¨\sqrt{3}£©^{4}•4}$-$\frac{1}{£¨\sqrt{3}£©^{2}•2}$+$\frac{1}{£¨\sqrt{3}£©^{5}•5}$-$\frac{1}{£¨\sqrt{3}£©^{3}•3}$
+¡+$\frac{1}{£¨\sqrt{3}£©^{n+1}£¨n+1£©}$-$\frac{1}{£¨\sqrt{3}£©^{n-1}£¨n-1£©}$+$\frac{1}{£¨\sqrt{3}£©^{n+2}£¨n+2£©}$-$\frac{1}{£¨\sqrt{3}£©^{n}n}$]
=$\frac{1}{2}$[$\frac{1}{£¨\sqrt{3}£©^{n+1}£¨n+1£©}$+$\frac{1}{£¨\sqrt{3}£©^{n+2}£¨n+2£©}$-$\frac{1}{\sqrt{3}}$-$\frac{1}{6}$]
£¾$\frac{1}{2}$£¨-$\frac{1}{\sqrt{3}}$-$\frac{1}{6}$£©=-$\frac{{2\sqrt{3}+1}}{12}$£®
¼´ÎªSn£¾-$\frac{{2\sqrt{3}+1}}{12}$£®
µãÆÀ ±¾Ì⿼²éµÈ²îÊýÁе͍ÒåºÍͨÏʽµÄÔËÓ㬿¼²é·½³Ì˼ÏëºÍÔËËãÄÜÁ¦£¬ÒÔ¼°ÁÑÏîÏàÏûÇóºÍ·½·¨£¬²»µÈʽµÄÐÔÖÊ£¬ÊôÓÚÖеµÌ⣮
| Äê¼¶ | ¸ßÖÐ¿Î³Ì | Äê¼¶ | ³õÖÐ¿Î³Ì |
| ¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | $\sqrt{2}$ | B£® | $\sqrt{3}$ | C£® | 2 | D£® | $\sqrt{6}$ |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
| Àí¿Æ | ÎÄ¿Æ | |
| ÄÐ | 13 | 10 |
| Å® | 7 | 20 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¹ú¼ÊѧУÓÅÑ¡ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com