·ÖÎö £¨1£©µ±n¡Ý2ʱ£¬an=Sn-Sn-1£¬ÍƵ¼³ö${S_n}{S_{n-1}}=\frac{1}{2}{S_n}-\frac{1}{2}{S_{n-1}}$£¬´Ó¶ø$\frac{1}{S_n}-\frac{1}{{{S_{n-1}}}}=2$£¬ÓÉ´ËÄÜÖ¤Ã÷$\{\frac{1}{S_n}\}$ÊǵȲîÊýÁУ¬´Ó¶øÄÜÇó³öSn£®
£¨2£©ÓÉ${b_n}=\frac{1}{£¨2n-1£©£¨2n+3£©}=\frac{1}{4}£¨\frac{1}{2n-1}-\frac{1}{2n+3}£©$£¬ÀûÓÃÁÑÏîÇóºÍ·¨ÄÜÇó³öTn£®
£¨3£©ÓÉ$£¨4{n^2}-4n+10£©•{S_n}=\frac{{4{n^2}-4n+10}}{2n-1}=\frac{{{{£¨2n-1£©}^2}+9}}{2n-1}$=$2n-1+\frac{9}{2n-1}$£¬·ÖnÎªÆæÊýºÍnΪżÊýÁ½ÖÖÇé¿ö½øÐÐÌÖÂÛ£¬ÄÜÇó³öʵÊýaµÄȡֵ·¶Î§£®
½â´ð Ö¤Ã÷£º£¨1£©µ±n¡Ý2ʱ£¬an=Sn-Sn-1£¬
ÓÉÒÑÖªÓÐ${S_n}^2=£¨{S_n}-{S_{n-1}}£©£¨{S_n}-\frac{1}{2}£©$£¬
¼´${S_n}{S_{n-1}}=\frac{1}{2}{S_n}-\frac{1}{2}{S_{n-1}}$£¬
¡à$\frac{1}{S_n}-\frac{1}{{{S_{n-1}}}}=2$£¬¡£¨3·Ö£©
¡ß$\frac{1}{{S}_{1}}$=$\frac{1}{{a}_{1}}$=1£¬¡à$\{\frac{1}{S_n}\}$ÊÇÊ×ÏîΪ1£¬¹«²îΪ2µÄµÈ²îÊýÁУ¬
¡à$\frac{1}{{S}_{n}}=1+£¨n-1£©¡Á2$=2n-1£¬
¡à${S_n}=\frac{1}{2n-1}$¡£¨4·Ö£©
½â£º£¨2£©¡ß${b_n}=\frac{1}{£¨2n-1£©£¨2n+3£©}=\frac{1}{4}£¨\frac{1}{2n-1}-\frac{1}{2n+3}£©$£¬
¡à${T_n}=\frac{1}{4}£¨1-\frac{1}{5}+\frac{1}{3}-\frac{1}{7}+\frac{1}{5}-\frac{1}{9}+\frac{1}{7}-\frac{1}{11}+¡+\frac{1}{2n-3}-\frac{1}{2n+1}+\frac{1}{2n-1}-\frac{1}{2n+3}£©$
=$\frac{1}{4}£¨\frac{4}{3}-\frac{1}{2n+1}-\frac{1}{2n+3}£©$
=$\frac{1}{3}-\frac{1}{4}£¨\frac{1}{2n+1}+\frac{1}{2n+3}£©$£®¡£¨8·Ö£©
£¨3£©¡ß$£¨4{n^2}-4n+10£©•{S_n}=\frac{{4{n^2}-4n+10}}{2n-1}=\frac{{{{£¨2n-1£©}^2}+9}}{2n-1}$=$2n-1+\frac{9}{2n-1}$£¬
¡àµ±nÎªÆæÊýʱ $-a£¼2n-1+\frac{9}{2n-1}$
Áît=2n-1£¬Ôò $y=t+\frac{9}{t}$£¬¡àt=5£¬¼´n=3ʱ£¬$-a£¼\frac{34}{5}$£¬¼´$a£¾-\frac{34}{5}$£¬
µ±nΪżÊýʱ£¬$a£¼2n-1+\frac{9}{2n-1}$£¬
ÓÖ$2n-1+\frac{9}{2n-1}¡Ý2\sqrt{9}=6$£¬
µ±n=2ʱȡ¡°=¡±£¬¡àa£¼6£¬
×ÛÉÏÌÖÂÛµÃ$-\frac{34}{5}£¼a£¼6$£®¡£¨12·Ö£©
µãÆÀ ±¾Ì⿼²éµÈ²îÊýÁеÄÖ¤Ã÷£¬¿¼²éÊýÁеÄǰnÏîºÍ¹«Ê½µÄÇ󷨣¬¿¼²éʵÊýµÄȡֵµÄÇ󷨣¬¿¼²éÍÆÀíÂÛÖ¤ÄÜÁ¦¡¢ÔËËãÇó½âÄÜÁ¦£¬¿¼²é»¯¹éÓëת»¯Ë¼Ïë¡¢º¯ÊýÓë·½³Ì˼Ï룬ÊÇÖеµÌ⣮
| 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£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
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£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | -$\frac{1}{5}$ | B£® | $\frac{1}{5}$ | C£® | -5 | D£® | 5 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | ³ä·Ö²»±ØÒªÌõ¼þ | B£® | ±ØÒª²»³ä·ÖÌõ¼þ | ||
| C£® | ³äÒªÌõ¼þ | D£® | ¼È²»³ä·ÖÒ²²»±ØÒªÌõ¼þ |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
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