高考复习方案2016届高考数学（理科浙江专用）二轮专题复习专题限时集训：（九）数列求和及数列的简单应用a word版含答案.doc

1、 1 a n n S n a 5 a 14 10 S 18 ( ) A 20 B 60 C 90 D 100 2 a n a 4 a 8 25 log 5 a 1 log 5 a 2 log 5 a 11 ( ) A 7 B 9 C 11 D 13 3 a n 4 15 55 ( ) A 10 B 20 C 30 D 40 4 S n a n n S 4 S 2 3 S 6 S 4 ( ) A 2 B. 7 3C. 3 10D 1 2 5 a n a 2 n 1 6a 2 n a n 1 a n a 1 2 a n n S n _ 6 a n n S n a 1 1 n 2 a n 2S n

2、 1 n S 2015 ( ) A 2015 B 2013 C 1008 D 1007 7 a n n S n S n S 2n a n b n 1 0 b n b n ( ) A b n n 1 B b n 2n 1 C b n n 1 D b n 2n 1 8 a n a 1 d n S n . y 1 2 a 1 x m (x 2) 2 y 2 1 x y d 0 1 S n 10 ( ) A. 9 10B. 10 11C. 8 9D 2 9 a n a n ( 1) n (2n 1)cos n 2 1(n N * ) n S n S 60 ( ) A 30 B 60 C 90 D 12

3、0 10 a n n S n S n 2a n 2 a n _ T n a 1 3a 2 (2n 1)a n T n _ 11 a n q n S n . n N * S 2n 50 n 16 a n P(a 2 14) Q(a 4 14) y x 45 x (1) a n n S n (2) b n 1 n a n n n 1 b n n T n . ( )A 1 C 解析 a n S 18 18 a 1 a 18 2 9(a 5 a 14 ) 90. 2 C a n a 4 a 8 a 1 a 11 a 2 a 10 a 3 a 9 a 5 a 7 a 2 6 25 log 5 a 1

4、log 5 a 2 log 5 a 11 log 5 (a 1 a 11 ) log 5 (a 2 a 10 ) log 5 (a 3 a 9 ) log 5 (a 4 a 8 ) log 5 (a 5 a 7 ) log 5 a 6 5log 5 25 log 5 5 11. 3 B 解析 a n 4 n n S n S S d 2 n 2n 40 n 20 20. 4 B 解析 a n q q 1 S 4 S 2 3 q 1 S n a 1 1 q n 1 q . S 4 S 2 3 1 q 4 1 q 2 1 q 2 3 q 2 2 S 6 S 4 1 q 6 1 q 4 1 8 1 4

5、 7 3 . 5 3 n 1 解析 a 2 n 1 6a 2 n a n 1 a n (a n 1 3a n )(a n 1 2a n ) 0 a n 0 a n 1 3a n a n 2 3 S n 2 1 3 n 1 3 3 n 1. 6 C 解析 n 2 a n 2S n 1 n a n 1 2S n n 1 a n 1 a n 1(n 2) a 1 1 S 2015 a 1 (a 2 a 3 ) (a 2014 a 2015 ) 1008. 7 B 解析 b n d(d 0) n T n T n T 2n k. b 1 1 n 1 2 n(n 1)d k 2n 1 2 2n 2n 1

6、d 2 (n 1)d 4k 2k(2n 1)d (4k 1)dn (2k 1)(2 d) 0. n (4k 1)d 0 (2k 1)(2 d) 0 d 2 k 1 4 . b n b n 2n 1. 8 B 解析 y 1 2 a 1 x m x y d 0 (2 0) x y d 0 a 1 2 d 2 a n n S n 2n n n 1 2 2 n(n 1) 1 S n 1 n n 1 1 n 1 n 1 1 S n 10 1 S 1 1 S 2 1 S 10 1 1 2 1 2 1 3 1 10 1 11 1 1 11 10 11 . 9 D 解析 n 4k 3(k N * ) a n

7、a 4k 3 1 n 4k 2(k N * ) a n a 4k 2 6 8k n 4k 1(k N * ) a n a 4k 1 1 n 4k(k N * ) a n a 4k 8k. a 4k 3 a 4k 2 a 4k 1 a 4k 8(k N * ) S 60 8 15 120. 10 2 n6 (2n 3)2 n 1 解析 n 2 a n S n S n 1 2a n 2a n 1 a n 2a n 1 n 1 a 1 S 1 2a 1 2 a 1 2. a n 2 n T n 2 3 2 2 (2n 1)2 n 2T n 2 2 3 2 3 (2n 1)2 n 1 T n 6 (2

8、n 3)2 n 1 . 11.( 0 1 解析 a n 0 q0. q 1 S 2n 0 q1 050 2 n 1 250 2 n 1 52 n 5 n N * S n n2 n 1 50 n 5. 16 解：(1) a 2 a 4 x 45 x 14 x 2 14x 45 0 a 2 a 4 14 a 2 a 4 45 a 2 5 a 4 9 a 2 9 a 4 5. a n a 2 5 a 4 9 d a 4 a 2 4 2 2 a n a 2 (n 2)d 5 (n 2) 2 2n 1 S n a 1 a n n 2 n(n 2) (2) (1) b n 1 n 2n 1 n n 1 (

9、 1) n 1 n 1 n 1 b n b n 1 ( 1) n 1 n 1 n 1 1 n 1 1 n 2 ( 1) n 1 n 1 n 2 . n T n (b 1 b 2 ) (b 3 b 4 ) (b n 1 b n ) ( 1) 1 1 1 3 ( 1) 3 1 3 1 5 ( 1) n 1 1 n 1 1 n 1 1 1 3 1 3 1 5 1 n 1 1 n 1 1 1 n 1 n n 1 . n T n n 1 n ( 1) n 1 n 1 n 1 1 1 n 1 n 1 n 1 1 1 n 1 n 2 n 1 . T n n n 1 n 2k k N * n 2 n 1 n 2k 1 k N * .