山东省百校大联考2024-2025学年高三上学期12月学情诊断数学试题（含解析）

绝密 ★ 启用前 a2 2n an 1 2(n 1) 1 2n 1, a2n 1 a 2 n 2 2(n 2) 1 2n 3, 2025 届高三上学期学情诊断 a2 a2 2 2 1 5, 数学答案及评分标准 3 22024.12 a2 a22 1 2 1 1 3, a2 a2 3 5 (2n 3) (2n 1) (n 1)(3 2n 1)将以上各式相加，得 n 1 n 2 1 ， 一、选择题：本题共 8 小题，每小题 5 分，共 40 分． 在每小题给出的四个选项中，只有一个选项是正确 2 a 1 a2 2 2 2的．请把正确的选项填涂在答题卡相应的位置上． 将 1 代入上式即得 n n ，且当 n 1时也成立，所以 an n ， 二、 又因为数列 an 为正项数列，所以 an n (n N ) ．····································································· 6分 1．A 2．C 3．B 4．B 5．D 6．C 7．A 8．C n n n 二、选择题：本题共 3 小题，每小题 6 分，共 18 分．在每小题给出的选项中，有多项符合题目要求．全 （2）由（1）可得bn ( 1) n 3 ，令 cn ( 1) n，其前 2n项和为T2n， 部选对得 6 分，部分选对的得部分分，选对但不全的得部分分，有选错的得 0 分． 则T2n 1 2 3 4 (2n 1) 2n n，··············································································9分 9．AC 10．BCD 11．ACD 31 32 32n 3(1 3 2n ) 3(32n 1) 32n 1 3 又因为 ，························································ 12分 1 3 2 2 三、填空题：本题共 3 小题，每小题 5 分，共 15 分． 32n 1 3 所以 S2n n ．··························································································· 13分2 1 π 16．[ 2, 6] 【解析】正弦定理+最值12． 13． 14． 4 3 ， 2 9 1 bcosC 3bsinC sin BcosC 3sin BsinC（ ）根据正弦定理， 1可化为 1，1分 a c sin A sinC 四、解答题：本题共 5 小题，共 77 分． 解答应写出文字说明、证明过程或演算步骤． sin BcosC 3sin BsinC sin A sinC 15．（1）【法一】构造常数列 sin BcosC 3sin BsinC sin(B C) sinC a2 2 2 2 2 2 2 2 2 2 sin BcosC 3sin BsinC sin BcosC sinC cosB sinC 由 n 1 an 2n 1 =(n+1) n (n N ) ， a1 1，可得 an 1 (n+1) an n a1 1 0 ， sinC( 3sin B cosB 1) 0. 2 故数列 an n2 2 2是恒为0的常数列，所以 an n ，······································································· 5分 ············································································································································· 5分 a C (0, π) π 又因为数列 为正项数列，所以 a n (n N ) .······································································· 6分 因为 ，所以 sinC 0，故有 3 sin B cosB 1 0，进而有 sin(B ) 1 ，因为 B (0, π) ， n n 6 2 【法二】累加法 π π所以 B ( , 5π) B π π π，故有 ，所以 B ．······························································· ... ...