15£®Ò»ÖÖ¼ÆËã×°Öã¬ÓÐÒ»Êý¾ÝÈë¿ÚAºÍÒ»¸öÔËËã³ö¿ÚB£¬°´ÕÕijÖÖÔËËã³ÌÐò£º
¢Ùµ±´ÓA¿ÚÊäÈë×ÔÈ»Êý1ʱ£¬´ÓB¿ÚµÃµ½$\frac{1}{3}$£¬¼ÇΪ$f£¨1£©=\frac{1}{3}$£»
¢Úµ±´ÓA¿ÚÊäÈë×ÔÈ»Êýn£¨n¡Ý2£©Ê±£¬ÔÚB¿ÚµÃµ½µÄ½á¹ûf£¨n£©ÊÇǰһ¸ö½á¹ûf£¨n-1£©µÄ$\frac{{2£¨{n-1}£©-1}}{{2£¨{n-1}£©+3}}$±¶£®
£¨1£©µ±´ÓA¿Ú·Ö±ðÊäÈë×ÔÈ»Êý2£¬3£¬4 Ê±£¬´ÓB¿Ú·Ö±ðµÃµ½Ê²Ã´Êý£¿²¢Çóf£¨n£©µÄ±í´ïʽ£»
£¨2£©¼ÇSnΪÊýÁÐ{f£¨n£©}µÄǰnÏîµÄºÍ£®µ±´ÓB¿ÚµÃµ½16112195µÄµ¹Êýʱ£¬Çó´Ëʱ¶ÔÓ¦µÄSnµÄÖµ£®

·ÖÎö £¨1£©¸ù¾Ýf£¨n£©=$\frac{2n-3}{2n+1}$f£¨n-1£©£¬ÒÀ´Î¼ÆËãf£¨2£©£¬f£¨3£©£¬f£¨4£©£¬¸ù¾Ý¹æÂɲÂÏëf£¨n£©£¬ÀûÓÃÊýѧ¹éÄÉ·¨Ö¤Ã÷£»
£¨2£©Áîf£¨n£©=$\frac{1}{16112195}$£¬¼ÆËãn£¬Ê¹ÓÃÁÐÏî·¨Çó³öSn£®

½â´ð ½â£º£¨1£©ÓÉÌâÒâµÃf£¨n£©=$\frac{2n-3}{2n+1}$f£¨n-1£©£¬f£¨1£©=$\frac{1}{3}$£¬
¡àf£¨2£©=$\frac{1}{5}$f£¨1£©=$\frac{1}{15}$£¬
f£¨3£©=$\frac{3}{7}$f£¨2£©=$\frac{1}{35}$£¬
f£¨4£©=$\frac{5}{9}$f£¨3£©=$\frac{1}{63}$£®
²Â²â£ºf£¨n£©=$\frac{1}{£¨2n-1£©£¨2n+1£©}$£®
ÏÔÈ»n=1ʱ£¬²ÂÏë³ÉÁ¢£¬
¼ÙÉèn=kʱ£¬²ÂÏë³ÉÁ¢£¬¼´f£¨k£©=$\frac{1}{£¨2k-1£©£¨2k+1£©}$£¬
Ôòn=k+1ʱ£¬f£¨k+1£©=$\frac{2k-1}{2k+3}$f£¨k£©=$\frac{1}{£¨2k+1£©£¨2k+3£©}$=$\frac{1}{[2£¨k+1£©-1][2£¨k+1£©+1]}$£¬
¡àµ±n=k+1ʱ£¬²ÂÏë³ÉÁ¢£¬
¡àf£¨n£©=$\frac{1}{£¨2n-1£©£¨2n+1£©}$£®
£¨2£©Áîf£¨n£©=$\frac{1}{£¨2n-1£©£¨2n+1£©}$=$\frac{1}{16112195}$µÃ4n2=16112196£¬¡àn=2007£¬
¡àSn=f£¨1£©+f£¨2£©+f£¨3£©+¡­+f£¨2007£©=$\frac{1}{1¡Á3}+\frac{1}{3¡Á5}$+$\frac{1}{5¡Á7}$+¡­+$\frac{1}{4013¡Á4015}$
=$\frac{1}{2}$£¨1-$\frac{1}{3}$£©+$\frac{1}{2}$£¨$\frac{1}{3}-\frac{1}{5}$£©+¡­+$\frac{1}{2}$£¨$\frac{1}{4013}$-$\frac{1}{4015}$£©
=$\frac{1}{2}$£¨1-$\frac{1}{3}+\frac{1}{3}$-$\frac{1}{5}$+¡­+¡­$\frac{1}{4013}$-$\frac{1}{4015}$£©
=$\frac{1}{2}$£¨1-$\frac{1}{4015}$£©=$\frac{2007}{4015}$£®

µãÆÀ ±¾Ì⿼²éÁËÊýÁÐͨÏʽµÄÇ󷨣¬ÊýÁÐÇóºÍ£¬ÊôÓÚÖеµÌ⣮

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

8£®½âÏÂÁв»µÈʽ£¨×飩
£¨1£©2x2-3x-5¡Ý£¨$\frac{1}{2}$£©x+2£»          
£¨2£©$\left\{\begin{array}{l}{\frac{2x+1}{x-3}£¾1}\\{{x}^{2}+x-20¡Ü0}\end{array}\right.$£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

6£®ÔÚ¡÷ABCÖУ¬Èô$£¨\;{a^2}+{c^2}-{b^2}£©tanB=\sqrt{3}$ac£¬Ôò½ÇB=£¨¡¡¡¡£©
A£®30¡ãB£®60¡ãC£®60¡ã»ò120¡ãD£®30¡ã»ò150¡ã

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

3£®ÏÂÁм¸ºÎÌåÊÇ×éºÏÌåµÄÊÇ£¨¡¡¡¡£©
A£®B£®C£®D£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

10£®Ä³Ð£¸ßÒ»Äê¼¶¾ÙÐÐÁËÒ»´ÎÊýѧ¾ºÈü£¬ÎªÁËÁ˽Ȿ´Î¾ºÈüѧÉúµÄ³É¼¨Çé¿ö£¬´ÓÖгéÈ¡Á˲¿·ÖѧÉúµÄ·ÖÊý£¨µÃ·ÖÈ¡ÕýÕûÊý£¬Âú·ÖΪ100·Ö£©×÷ΪÑù±¾£¨Ñù±¾ÈÝÁ¿Îªn£©½øÐÐͳ¼Æ£¬°´ÕÕ[50£¬60£©£¬[60£¬70£©£¬[70£¬80£©£¬[80£¬90£©£¬[90£¬100]µÄ·Ö×é×÷³öƵÂÊ·Ö²¼Ö±·½Í¼£¬²¢×÷³öÑù±¾·ÖÊýµÄ¾¥Ò¶Í¼£¨Í¼ÖнöÁгöÁ˵÷ÖÔÚ[50£¬60£©£¬[90£¬100]µÄÊý¾Ý£©£®

£¨1£©ÇóÑù±¾ÈÝÁ¿nºÍƵÂÊ·Ö²¼Ö±·½Í¼ÖеÄx£¬yµÄÖµ£»
£¨2£©¹À¼Æ±¾´Î¾ºÈüѧÉú³É¼¨µÄÖÐλÊýºÍƽ¾ù·Ö£»
£¨3£©ÔÚѡȡµÄÑù±¾ÖУ¬´Ó¾ºÈü³É¼¨ÔÚ80·ÖÒÔÉÏ£¨º¬80·Ö£©µÄѧÉúÖÐËæ»ú³éÈ¡2ÃûѧÉú£¬ÇóËù³éÈ¡µÄ2ÃûѧÉúÖÐÖÁÉÙÓÐÒ»È˵÷ÖÔÚ[90£¬100]ÄڵĸÅÂÊ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

20£®ÈôÒ»ÉÈÐεÄÔ²ÐĽÇΪ3»¡¶È£¬ÇÒ´ËÉÈÐÎÖܳ¤Îª5£¬Ôò´ËÉÈÐεÄÃæ»ýS=$\frac{3}{2}$£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

7£®ÒÑÖªÖ±Ïßx=my+1¹ýÅ×ÎïÏßC£ºy2=2px£¨p£¾0£©µÄ½¹µãFÇÒÓëÅ×ÎïÏßÏཻÓÚÁ½µãM£¨x1£¬y1£©£¬N£¨x2£¬y2£©£¬×ÔM£¬NÏò×¼ÏßL×÷´¹Ïߣ¬´¹×ã·Ö±ðΪM1£¬N1£®
£¨¢ñ£©ÇóÅ×ÎïÏßCµÄ·½³Ì£»
£¨¢ò£©Ö¤Ã÷£ºÎÞÂÛmÈ¡ºÎʵÊýʱ£¬y1y2£¬x1x2¶¼ÊǶ¨Öµ£»
£¨¢ó£©¼Ç¡÷FMM1£¬¡÷FM1N1£¬¡÷FNN1µÄÃæ»ý·Ö±ðΪS1£¬S2£¬S3£¬ÊÔÅжÏ$S_2^2=4{S_1}{S_3}$ÊÇ·ñ³ÉÁ¢£¬²¢Ö¤Ã÷ÄãµÄ½áÂÛ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

4£®½Ç¦ÁµÄÖÕ±ßÉÏÓÐÒ»µãM£¨-2£¬4£©£¬Ôòtan¦Á=-2£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

5£®µØÇò³àµÀµÄ°ë¾¶Îª6370km£¬Ôò³àµÀÉÏ1»¡¶È½ÇËù¶ÔµÄÔ²»¡³¤Îª6370km£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸