数列经典题目

高考真题、典型问题

题目

(2021全国乙卷)                                                                                                                         Sn为数列{an}的前n项和,bn为数列{Sn}的前n项积,已知2Sn+1bn=2.                      (1)证明:数列{bn}是等差数列;(2){an}的通项公式.                                                     (2021 全国乙卷)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\记 S_{n} 为数列 \left\{a_{n}\right\} 的前 n 项和, b_{n} 为数列 \left\{S_{n}\right\} 的前 n 项积,已知 \frac{2}{S_{n}}+\frac{1}{b_{n}}=2 .~~~~~~~~~~~~~~~~~~~~~~\\ (1)证明: 数列 \left\{b_{n}\right\} 是等差数列; (2)求 \left\{a_{n}\right\} 的通项公式.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案

(1){bn}是以32为首项,12为公差的等差数列.(2)an={32,n=1,1n(n+1),n2.                 提示:Sn=bnbn1.                                                                                                                      (1) \left\{b_{n}\right\} 是以 \frac{3}{2} 为首项, \frac{1}{2} 为公差的等差数列. (2) a_{n}=\left\{\begin{array}{l}\displaystyle\frac{3}{2}, n=1, \\ -\displaystyle\frac{1}{n(n+1)}, n \geqslant 2 .\end{array}\right.~~~~~~~~~~~~~~~~~\\ 提示:S_n=\frac{b_n}{b_{n-1}}.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

(2023新高考1)                                                                                                                       设等差数列{an}的公差为d,d>1.bn=n2+nan,Sn,Tn分别为数列{an},{bn}n项和.                                                                                                                                     (1)3a2=3a1+a3,S3+T3=21,{an}的通项公式;                                                     (2){bn}为等差数列,S99T99=99,d.                                                                      (2023新高考 1 卷) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\设等差数列 \left\{a_{n}\right\} 的公差为 d , 且 d>1 . 令 b_{n}=\frac{n^{2}+n}{a_{n}} , 记 S_{n}, T_{n} 分别为数列 \left\{a_{n}\right\},\left\{b_{n}\right\} 的\\前 n 项和.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (1)若 3 a_{2}=3 a_{1}+a_{3}, S_{3}+T_{3}=21 , 求 \left\{a_{n}\right\} 的通项公式;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (2)若 \left\{b_{n}\right\} 为等差数列, 且 S_{99}-T_{99}=99 , 求 d .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案

(1)an=3n(2)d=5150.                                                                                                           (1) a_{n}=3 n;(2) d=\frac{51}{50} .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

图中的数阵满足:每一行从左到右成等差数列,每一列从上到下成等比数列,且          公比均为实数q,  a1,1>0,a1,3=5,a2,2=6,a4,22=a7,5.                                                 a1,1a1,2a1,3a1,na2,1a2,2a2,3a2,na3,1a3,2a3,3a3,nan,1an,2an,3an,n(1)bn=an,n,求数列{bn}的通项公式;                                                                            (2)Sn=a1,1+a2,1++an,1,是否存在实数λ,使an,1λSn恒成立,若存在,          求出λ的所有值,若不存在,说明理由.                                                                               图中的数阵满足: 每一行从左到右成等差数列, 每一列从上到下成等比数列, 且~~~~~~~~~~\\公比均为实数 q, ~~a_{1,1}>0, a_{1,3}=5, a_{2,2}=-6,a_{4,2}^{2}=a_{7,5}.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ \begin{matrix} &a_{1,1} &a_{1,2} &a_{1,3} &\cdots &a_{1,n}&\cdots\\ &a_{2,1} &a_{2,2} &a_{2,3} &\cdots &a_{2,n}&\cdots\\ &a_{3,1} &a_{3,2} &a_{3,3} &\cdots &a_{3,n}&\cdots\\ &\cdots & & & & \\ &a_{n,1} &a_{n,2} &a_{n,3} &\cdots &a_{n,n}&\cdots\\ &\cdots & & & & \end{matrix}\\ (1)设b_n=a_{n,n},求数列\left \{b_n \right \} 的通项公式;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (2)设S_n=a_{1,1}+a_{2,1}+\cdots+a_{n,1},是否存在实数\lambda ,使a_{n,1}\le\lambda S_n恒成立,若存在,~~~~~~~~~~\\求出\lambda 的所有值,若不存在,说明理由.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案

答案:λ=32.                                                                                                                             答案:\lambda = \frac{3}{2}.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

n2(n4)个正数,排成n×n矩阵(nn列的数表):                                                   (a11a12a1na21a22a2nan1an2ann)其中每一行的数成等差数列,每一列的数成等比数列,并且所有的公比都相等,已知a24=1,a42=18,a43=316(1)求公比q;(2)k表示a4k;(3)a11+a22++ann.          有n^2(n\ge 4)个正数,排成n\times n矩阵(n行n列的数表):~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ \begin{pmatrix} a_{11}& a_{12}& \cdots & a_{1n} \\ a_{21}& a_{22}& \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1}& a_{n2}& \cdots & a_{nn} \end{pmatrix} \\ 其中每一行的数成等差数列,每一列的数成等比数列,并且所有的公比都相等,已知\\a_{24}=1,a_{42}=\frac{1}{8} ,a_{43}=\frac{3}{16} (1)求公比q;(2)用k表示a_{4k};(3)求a_{11}+a_{22}+\cdots +a_{nn}.~~~~~~~~~~
答案

答案:(1)q=12;(2)a4k=k16;(3)S=2n+22n                                                                   答案:(1)q=\frac{1}{2};(2)a_{4k}=\frac{k}{16};(3)S=2-\frac{n+2}{2^n}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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