裂项相消法 · 历年高考数学真题与解析

本页汇总 高考真题检索 的「裂项相消法」高考数学真题共 12 道,覆盖 2011–2017 年,最常出题型为 解答题;含完整答案与解析。

12questions
Past questions included
2011–2017
Years covered
区分题为主
Overall difficulty
Free-response
Most common question type
Practice This Topic Do 5 reliable past questions first, then see the answers and explanations after you submit
常用解题方法化归与转化整体代换构造法
常见易错点数列下标错位漏解符号错误
Core competencies应用

Past Question List

2017 National Gaokao Fill in the blank Differentiating Question 15 2017_新课标 II 卷 (2017·理)

15.(5分)等差数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和为 $S_{n}, a_{3}=3, S_{4}=10$ ,则 $\sum_{k=1}^{n} \frac{1}{S_{k}}=-\frac{2 n}{\underline{n+1}}$-

2017 National Gaokao Free response Differentiating Question 17 2017_新课标 III 卷 (2017·文)

17.(12分)设数列 $\left\{a_{n}\right\}$ 满足 $a_{1}+3 a_{2}+\ldots+(2 n-1) a_{n}=2 n$ .
(1)求 $\left\{a_{n}\right\}$ 的通项公式;
(2)求数列 $\left\{\frac{{ }^{a} n}{2 n+1}\right\}$ 的前 $n$ 项和.

2015 Jiangsu Gaokao Fill in the blank Differentiating Question 11 2015_江苏卷 (2015)

11.(5分)(2015•江苏)设数列 $\left\{a_{n}\right\}$ 满足 $a_{1}=1$ ,且 $a_{n+1}-a_{n}=n+1 \quad\left(n \in N^{*}\right)$ ,则数列 $\left\{\frac{1}{a_{n}}\right\}$的前 10 项的和为 $\_\_\_\_$。

2015 National Gaokao Free response Differentiating Question 17 2015_新课标 I 卷 (2015·理)

17.(12分) $\mathrm{S}_{\mathrm{n}}$ 为数列 $\left\{\mathrm{a}_{\mathrm{n}}\right\}$ 的前 n 项和,已知 $\mathrm{a}_{\mathrm{n}}>0, \mathrm{a}_{\mathrm{n}}{ }^{2}+2 \mathrm{a}_{\mathrm{n}}=4 \mathrm{~S}_{\mathrm{n}}+3$
(I)求 $\left\{a_{n}\right\}$ 的通项公式:
(II)设 $b_{n}=\frac{1}{a_{n} a_{n+1}}$ ,求数列 $\left\{b_{n}\right\}$ 的前 $n$ 项和。

2015 National Gaokao Free response Differentiating Question 3 2015_退役省自主命题 (2015·文)

18.已知数列 $\left\{a_{n}\right\}$ 是递增的等比数列,且 $a_{1}+a_{4}=9, a_{2} a_{3}=8$ .
(I)求数列 $\left\{a_{n}\right\}$ 的通项公式;

(II)设 $S_{n}$ 为数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和,$b_{n}=\frac{a_{n+1}}{S_{n} S_{n+1}}$ ,求数列 $\left\{b_{n}\right\}$ 的前 $n$ 项和 $T_{n}$ .

2014 National Gaokao Free response Differentiating Question 17 2014_退役省自主命题 (2014·文)

19.(本小题满分 14 分)

设各项均为正数的数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和为 $S_{n}$ ,且 $S_{n}$ 满足 $S_{n}^{2}-\left(n^{2}+n-3\right) S_{n}-3\left(n^{2}+n\right)=0, n \in N^{*}$ .
(1)求 $a_{1}$ 的值;
(2)求数列 $\left\{a_{n}\right\}$ 的通项公式;
(3)证明:对一切正整数 $n$ ,有 $\frac{1}{a_{1}\left(a_{1}+1\right)}+\frac{1}{a_{2}\left(a_{2}+1\right)}+\cdots \frac{1}{a_{n}\left(a_{n}+1\right)}<\frac{1}{3}$ .

2012 National Gaokao Single choice Differentiating Question 5 2012_大纲版 (2012·理)

5.(5分)已知等差数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和为 $S_{n}, a_{5}=5, S_{5}=15$ ,则数列 $\left\{\frac{1}{a_{n} a_{n+1}}\right\}$的前 100 项和为( )

A. $\frac{100}{101}$
B. $\frac{99}{101}$
C. $\frac{99}{100}$
D. $\frac{101}{100}$
2011 National Gaokao Free response Differentiating Question 20 2011_大纲版 (2011·理)

20.(12分)设数列 $\left\{a_{n}\right\}$ 满足 $a_{1}=0$ 且 $\frac{1}{1-a_{n+1}}-\frac{1}{1-a_{n}}=1$ .
(I)求 $\left\{a_{n}\right\}$ 的通项公式;
(II)设 $\mathrm{b}_{\mathrm{n}}=\frac{1-\sqrt{\mathrm{a}_{\mathrm{n}+1}}}{\sqrt{\mathrm{n}}}$ ,记 $\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{b}_{\mathrm{k}}$ ,证明: $\mathrm{S}_{\mathrm{n}}<1$ .

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