2024学生版大二轮数学新高考提高版（京津琼鲁辽粤冀鄂湘渝闽苏浙黑吉晋皖云豫新甘贵赣桂）专题三　第2讲　数列求和及其综合应用50

第2讲数列求和及其综合应用[考情分析]1.数列求和重点考查分组转化、错位相减、裂项相消三种求和方法.2.数列的综合问题，一般以等差数列、等比数列为背景，与函数、不等式相结合，考查最值、范围以及证明不等式等.3.主要以选择题、填空题及解答题的形式出现，难度中等．考点一数列求和核心提炼1．裂项相消法就是把数列的每一项分解，使得相加后项与项之间能够相互抵消，但在抵消的过程中，有的是相邻项抵消，有的是间隔项抵消．常见的裂项方式有：eq\f(1,nn＋k)＝eq\f(1,k)eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,n)－\f(1,n＋k)))；eq\f(1,4n2－1)＝eq\f(1,2)eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,2n－1)－\f(1,2n＋1))).2．错位相减法求和，主要用于求{anbn}的前n项和，其中{an}，{bn}分别为等差数列和等比数列．考向1分组转化法例1(2023·枣庄模拟)已知数列{an}的首项a1＝3，且满足an＋1＋2an＝2n＋2.(1)证明：{an－2n}为等比数列；________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(2)已知bn＝eq\b\lc\{\rc\(\a\vs4\al\co1(an，n为奇数，,log2an，n为偶数，))Tn为{bn}的前n项和，求T10.________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________考向2裂项相消法例2(2023·沈阳质检)设n∈N*，向量eq\o(AB,\s\up6(→))＝(n－1,1)，eq\o(AC,\s\up6(→))＝(n－1,4n－1)，an＝eq\o(AB,\s\up6(→))·eq\o(AC,\s\up6(→)).(1)令bn＝an＋1－an，求证：数列{bn}为等差数列；________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(2)求证：eq\f(1,a1)＋eq\f(1,a2)＋…＋eq\f(1,an)<eq\f(3,4).________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________考向3错位相减法例3(2023·全国甲卷)记Sn为数列{an}的前n项和，已知a2＝1,2Sn＝nan.(1)求{an}的通项公式；(2)求数列eq\b\lc\{\rc\}(\a\vs4\al\co1(\f(an＋1,2n)))的前n项和Tn.________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________规律方法(1)分组转化法求和的关键是将数列通项转化为若干个可求和的数列通项的和或差．(2)裂项相消法的基本思路是将通项拆分，可以产生相互抵消的项．(3)用错位相减法求和时，应注意：①等比数列的公比为负数的情形；②在写出“Sn”和“qSn”的表达式时应特别注意将两式“错项对齐”，以便准确写出“Sn－qSn”的表达式．跟踪演练1(1)(2023·淮南模拟)已知数列{an}满足an＋1－an＝2n，且a1＝1.①求数列{an}的通项公式；②设bn＝eq\f(an＋1,anan＋1)，求数列{bn}的前n项和Tn.________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(2)(2023·浙江省强基联盟模拟)已知a1＝1，{an＋1}是公比为2的等比数列，{bn}为正项数列，b1＝1，当n≥2时，(2n－3)bn＝(2n－1)bn－1.①求数列{an}，{bn}的通项公式；②记cn＝an·bn.求数列{cn}的前n项和Tn.________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________考点二数列的综合问题核心提炼数列与函数、不等式，以及数列新定义的综合问题，是高考命题的一个方向，考查逻辑推理、数学运算、数学建模等核心素养．解决此类问题，一是把数列看成特殊的函数，利用函数的图象、性质求解；二是将新数列问题转化为等差或等比数列，利用特殊数列的概念、公式、性质，结合不等式的相关知识求解．例4(1)分形的数学之美，是以简单的基本图形，凝聚扩散，重复累加，以迭代的方式而形成的美丽的图案．自然界中存在着许多令人震撼的天然分形图案，如鹦鹉螺的壳、蕨类植物的叶子、孔雀的羽毛、菠萝等．如图所示，为正方形经过多次自相似迭代形成的分形图形，且相邻的两个正方形的对应边所成的角为15°.若从外往里最大的正方形边长为9，则第