Question

3£®¼ºÖªÊýÁÐ{a_n}Óë{b_n}¶¼ÊǵȲîÊýÁУ¬ÇÒ$\underset{lim}{n¡ú¡Þ}$$\frac{{a}_{n}}{{b}_{n}}$=3£¬Ôò$\underset{lim}{n¡ú¡Þ}$$\frac{{a}_{1}+{a}_{2}+¡­+{a}_{2n}}{{b}_{1}+{b}_{2}+¡­+{b}_{3n}}$µÄֵΪ£¨¡¡¡¡£©

• A£® $\frac{9}{4}$
• B£® $\frac{4}{3}$
• C£® $\frac{4}{3}$»ò2
• D£® $\frac{4}{3}$»ò$\frac{9}{4}$

Answer 1

·ÖÎö ÉèÊýÁÐ{a_n}Óë{b_n}µÄ¹«²î·Ö±ðÊÇa£¬b£¬ÔËÓõȲîÊýÁеÄͨÏʽ£¬ÒÔ¼°ÊýÁм«Ï޵Ĺ«Ê½¿ÉµÃa=3b£¬ÔÙÓɵȲîÊýÁеÄÇóºÍ¹«Ê½£¬¿ÉµÃËùÇó¼«ÏÞ£®

½â´ð ½â£ºÉèÊýÁÐ{a_n}Óë{b_n}µÄ¹«²î·Ö±ðÊÇa£¬b£¬
ÓÉ$\underset{lim}{n¡ú¡Þ}$$\frac{{a}_{n}}{{b}_{n}}$=3£¬¿ÉµÃ$\underset{lim}{n¡ú¡Þ}$$\frac{{a}_{1}+a£¨n-1£©}{{b}_{1}+b£¨n-1£©}$=3£¬
¼´Îª$\frac{a}{b}$=3£¬
ÓÖ$\underset{lim}{n¡ú¡Þ}$$\frac{{a}_{1}+{a}_{2}+¡­+{a}_{2n}}{{b}_{1}+{b}_{2}+¡­+{b}_{3n}}$=$\underset{lim}{n¡ú¡Þ}$$\frac{2n{a}_{1}+\frac{2n£¨2n-1£©}{2}a}{3n{b}_{1}+\frac{3n£¨3n-1£©}{2}b}$
=$\underset{lim}{n¡ú¡Þ}$$\frac{\frac{2{a}_{1}-a}{n}+2a}{\frac{6{b}_{1}-3b}{2n}+\frac{9b}{2}}$=$\frac{4a}{9b}$=$\frac{4}{3}$£®
¹ÊÑ¡£ºB£®

µãÆÀ ±¾Ì⿼²éÊýÁм«ÏÞµÄÇ󷨣¬¿¼²éµÈ²îÊýÁеÄͨÏʽºÍÇóºÍ¹«Ê½µÄÔËÓ㬻¯¼òÕûÀíµÄÔËËãÄÜÁ¦£¬ÊôÓÚ»ù´¡Ì⣮