Question

£¨2008•ÇàÆÖÇøһģ£©ÒÑÖªf£¨x£©=log₂x£¬Èô2£¬f£¨a₁£©£¬f£¨a₂£©£¬f£¨a₃£©£¬¡­£¬f£¨a_n£©£¬2n+4£¬¡­£¨n¡ÊN^*£©³ÉµÈ²îÊýÁУ®
£¨1£©ÇóÊýÁÐ{a_n}£¨n¡ÊN^*£©µÄͨÏʽ£»
£¨2£©Éèg£¨k£©ÊDz»µÈʽlog2x+log2(3

ak

-x)¡Ý2k+3(k¡ÊN*)ÕûÊý½âµÄ¸öÊý£¬Çóg£¨k£©£»
£¨3£©ÔÚ£¨2£©µÄÌõ¼þÏ£¬ÊÔÇóÒ»¸öÊýÁÐ{b_n}£¬Ê¹µÃ

lim

n¡ú¡Þ

[

1

g(1)g(2)

b1+

1

g(2)g(3)

b2+¡­

1

g(n)g(n+1)

bn]=

1

5

£®

Answer 1

·ÖÎö£º£¨1£©ÏÈŪÇåÊýÁеÄÏîÊý£¬È»ºó¸ù¾ÝµÈ²îÊýÁеÄͨÏʽÇó³ö¹«²îd£¬´Ó¶øÇó³öf£¨a_n£©µÄÖµ£¬¼´¿ÉÇó³öÊýÁÐ{a_n}£¨n¡ÊN^*£©µÄͨÏʽ£»
£¨2£©½«a_k´úÈë²»µÈʽ£¬È»ºó¸ù¾Ý¶ÔÊýµÄÔËËãÐÔÖʽøÐл¯¼ò±äÐΣ¬È»ºóÒòʽ·Ö½âµÃ£¨x-2^k+1£©£¨x-2•2^k+1£©¡Ü0£¬´Ó¶øÇó³öxµÄ·¶Î§£¬¼´¿ÉÇó³ög£¨k£©£»
£¨3£©½«

1

g(n)g(n+1)

½øÐÐÁÑÏîµÃ

1

g(n)g(n+1)

=

1

2n+1

(

1

2n+1+1

-

1

2n+2+1

)£¬¿ÉÈ¡b_n=2ⁿ⁺¹£¬È»ºóÑéÖ¤

lim

n¡ú¡Þ

[

1

g(1)g(2)

b1+

1

g(2)g(3)

b2+¡­

1

g(n)g(n+1)

bn]=

1

5

ÊÇ·ñ³ÉÁ¢£®

½â´ð£º½â£º£¨1£©2n+4=2+£¨n+1£©d£¬
¡àd=2    f£¨a_n£©=2+£¨n+1-1£©•2=2£¨n+1£©
¼´log₂a_n=2n+2£¬
¡àa_n=2²ⁿ⁺²
£¨2£©log2(-x2+3

22(k+1)

x)¡Ý2k+3£¬
¡à-x2+3

22(k+1)

x¡Ý22k+3£¬
µÃ£¬x²-3•2^k+1x+2^{2£¨k+1£©+1}¡Ü0£¬¼´x²-3•2^k+1x+2•£¨2^k+1£©²¡Ü0£¬
¡à£¨x-2^k+1£©£¨x-2•2^k+1£©¡Ü0£¬
¡à2^k+1¡Üx¡Ü2•2^k+1
Ôòg£¨k£©=2^k+1+1
£¨3£©

1

g(n)g(n+1)

=

1

(2n+1+1)(2n+2+1)

=

1

2n+1

(

1

2n+1+1

-

1

2n+2+1

)£¬
È¡b_n=2ⁿ⁺¹£¬
Ôò

1

g(n)g(n+1)

bn=

1

(2n+1+1)(2n+2+1)

bn=

1

2n+1+1

-

1

2n+2+1

lim

n¡ú¡Þ

[

1

g(1)g(2)

b1+

1

g(2)g(3)

b2+¡­

1

g(n)g(n+1)

bn]=

lim

n¡ú¡Þ

(

1

5

-

1

2n+2+1

)=

1

5

£®
¡àb_n=2ⁿ⁺¹

µãÆÀ£º±¾ÌâÖ÷Òª¿¼²éÁËÊýÁÐÓë²»µÈʽµÄ×ÛºÏÔËÓã¬Í¬Ê±¿¼²éÁËÁÑÏîÇóºÍ·¨ºÍ¼ÆËãÄÜÁ¦£¬ÊôÓÚÖеµÌ⣮