第1页共7页三明市2025年普通高中高三毕业班质量检测数学参考答案及评分细则评分说明：1．本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分标准制定相应的评分细则．2．对计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半；如果后继部分的解答有较严重的错误，就不再给分．3．解答右端所注分数，表示考生正确做到这一步应得的累加分数．4．只给整数分数．选择题和填空题不给中间分．一、选择题：本大题考查基础知识和基本运算．每小题5分，满分40分．1．C2．D3．B4．A5．B6．A7．A8．D二、选择题：本大题考查基础知识和基本运算．每小题6分，满分18分．全部选对的得6分，部分选对的得部分分，有选错的得0分．9．AC10．ACD11．BCD三、填空题：本大题考查基础知识和基本运算．每小题5分，满分15分．12．513．3314．10,4四、解答题：本大题共5小题，共77分．解答应写出文字说明、证明过程或演算步骤．15.（1）证明：因为nnnnnaa112)32(352，所以11352nnnaa，············································································1分）（nnnnnnaaa31236231111，··························································3分因为311a，所以0311a,031na,······················································4分所以2313111nnnnaa··················································································5分所以数列}31{nna为等比数列是首项为311a，公比为2的等比数列.···············6分（2）由（1）可得数列}31{nna为首项为2，公比为2的等比数列,所以nnna231,即nnna312································································7分所以233121231131312122111nnnnnS)(）（，（注：233123211nnnS)(应得分）··········································································································9分因为数列}{nS为单调递增数列，······························································10分{#{QQABBQYswggwwhSACB5KQ0GyCgsQkIGiJYoEgRAUOAxigJNIBKA=}#}第2页共7页当20251024233121299109)(Sn时，，当10n时，202523)31(212101110S，··············································12分所以，使不等式2025nS成立的n的最小值为10.······································13分16．解：（1）设A“一次回答问题，AI软件答对问题”，B“选出语文问题让AI回答”，（注：若设事件合理则得分）························1分依题意,53)(,52)(BPBP,·····································································2分43)|(BAP,21)|(BAP,·······································································3分所以5321534352)|()()|()()(...