Question

7£®Èçͼ£¬ÓÉÈô¸ÉÔ²µã×é³ÉÈçÈý½ÇÐεÄͼÐΣ¬Ã¿Ìõ±ß£¨°üÀ¨Á½¸ö¶Ëµã£©ÓÐn£¨n£¾1£¬n¡ÊN£©¸öµã£¬Ã¿¸öͼÐÎ×ܵĵãÊý¼ÇΪa_n£¬Ôò$\frac{9}{{{a_2}{a_3}}}+\frac{9}{{{a_3}{a_4}}}+\frac{9}{{{a_4}{a_5}}}+¡­+\frac{9}{{{a_{2014}}{a_{2015}}}}$=£¨¡¡¡¡£©

• A£® $\frac{2014}{2015}$
• B£® $\frac{2013}{2014}$
• C£® $\frac{3}{2015}$
• D£® $\frac{9}{2015}$

Answer 1

·ÖÎö ¸ù¾ÝͼÏóµÄ¹æÂɿɵóöͨÏʽa_n£¬¸ù¾ÝÊýÁÐ{$\frac{9}{{a}_{n}{a}_{n+1}}$}µÄÌصã¿ÉÓÃÁÐÏî·¨Çó³ö$\frac{9}{{a}_{2}{a}_{3}}+\frac{9}{{a}_{3}{a}_{4}}+\frac{9}{{a}_{4}{a}_{5}}+¡­+\frac{9}{{a}_{n}{a}_{n+1}}$=$\frac{n-1}{n}$£¬½«n=2014´úÈë¿ÉµÃ´ð°¸£®

½â´ð ½â£ºÃ¿¸ö±ßÓÐn¸öµã£¬°Ñÿ¸ö±ßµÄµãÊýÏà¼ÓµÃ3n£¬ÕâÑù½ÇÉϵĵãÊý±»Öظ´¼ÆËãÁËÒ»´Î£¬
¹ÊµÚn¸öͼÐεĵãÊýΪ3n-3£¬¼´a_n=3n-3£¬
ÁîS_n=$\frac{9}{{a}_{2}{a}_{3}}+\frac{9}{{a}_{3}{a}_{4}}+\frac{9}{{a}_{4}{a}_{5}}+¡­+\frac{9}{{a}_{n}{a}_{n+1}}$=$\frac{1}{1¡Á2}$+$\frac{1}{2¡Á3}$+¡­+$\frac{1}{£¨n-1£©n}$=1-$\frac{1}{2}$+$\frac{1}{2}$-$\frac{1}{3}$+¡­+$\frac{1}{n-1}$-$\frac{1}{n}$=1-$\frac{1}{n}$=$\frac{n-1}{n}$£¬
¡à$\frac{9}{{{a_2}{a_3}}}+\frac{9}{{{a_3}{a_4}}}+\frac{9}{{{a_4}{a_5}}}+¡­+\frac{9}{{{a_{2014}}{a_{2015}}}}$=$\frac{2013}{2014}$£¬
¹ÊÑ¡£ºB

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éµÈ²îÊýÁеÄͨÏʽºÍÇóºÍÎÊÌ⣮Êô»ù´¡Ì⣮