【答案】
分析:根據(jù)題意,數(shù)列{log
2(a
n-1)}(n∈N
*)為等差數(shù)列,設(shè)其公差為d,則log
2(a
n-1)-log
2(a
n-1-1)=d,由對(duì)數(shù)的運(yùn)算性質(zhì)可得,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212948758082401/SYS201310232129487580824002_DA/0.png)
=2
d,又由a
1=3,a
2=5,可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212948758082401/SYS201310232129487580824002_DA/1.png)
=2,則可得{a
n-1}是以a
1-1=2為首項(xiàng),公比為2的等比數(shù)列,進(jìn)而可得a
n=2
n+1,結(jié)合題意有a
n-a
n-1=2
n-2
n-1=2
n-1,代入可得答案.
解答:解:數(shù)列{log
2(a
n-1)}(n∈N
*)為等差數(shù)列,
設(shè)其公差為d,則log
2(a
n-1)-log
2(a
n-1-1)=d,
即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212948758082401/SYS201310232129487580824002_DA/2.png)
=2
d,又由a
1=3,a
2=5,
則d=1,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212948758082401/SYS201310232129487580824002_DA/3.png)
=2,
{a
n-1}是以a
1-1=2為首項(xiàng),公比為2的等比數(shù)列,
進(jìn)而可得,a
n-1=2
n,則a
n=2
n+1,
故a
n-a
n-1=2
n-2
n-1=2
n-1,
則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212948758082401/SYS201310232129487580824002_DA/4.png)
(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212948758082401/SYS201310232129487580824002_DA/5.png)
+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212948758082401/SYS201310232129487580824002_DA/6.png)
+…+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212948758082401/SYS201310232129487580824002_DA/7.png)
)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212948758082401/SYS201310232129487580824002_DA/8.png)
(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212948758082401/SYS201310232129487580824002_DA/9.png)
+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212948758082401/SYS201310232129487580824002_DA/10.png)
+…+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212948758082401/SYS201310232129487580824002_DA/11.png)
)=1,
故選C.
點(diǎn)評(píng):本題考查等差、等比數(shù)列的性質(zhì)與極限的運(yùn)算,注意與對(duì)數(shù)函數(shù)或指數(shù)函數(shù)的結(jié)合運(yùn)用時(shí),往往同時(shí)涉及等比、等差數(shù)列的性質(zhì),是一個(gè)難點(diǎn).