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

这是一份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，共5页。
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,nn＋k)＝eq \f(1,k)eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(1,n)－\f(1,n＋k)))；
eq \f(1,4n2－1)＝eq \f(1,2)eq \b\lc\(\rc\)(\a\vs4\al\c1(\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；
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(2)已知bn＝eq \b\lc\{\rc\ (\a\vs4\al\c1(an，n为奇数，,lg2an，n为偶数，))Tn为{bn}的前n项和，求T10.
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考向2 裂项相消法
例2 (2023·沈阳质检)设n∈N*，向量eq \(AB,\s\up6(→))＝(n－1,1)，eq \(AC,\s\up6(→))＝(n－1,4n－1)，an＝eq \(AB,\s\up6(→))·eq \(AC,\s\up6(→)).
(1)令bn＝an＋1－an，求证：数列{bn}为等差数列；
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(2)求证：eq \f(1,a1)＋eq \f(1,a2)＋…＋eq \f(1,an)