knowledge-vault/work/client-projects/高新教育/文枢/数学/试题16/题目.md

题目描述

原始文本

16. ($15$ 分)
已知等差数列 $\{a_n\}$ 满足 $a_1=1, a_{n+1}a_n=4n^2+\lambda$ ($\lambda$ 为常数).
$(1)$ 求 $\lambda$ 的值，并求 $\{a_n\}$ 的通项公式；
$(2)$ 求数列 $\left\{\dfrac{1}{a_{n+1}a_n}\right\}$ 的前 $n$ 项和 $S_n$.

答案

### 16. 命题透析

本题考查等差数列的定义及用裂项相消法求数列的前 $n$ 项和.

**解析**

(1) 设 $\{a_n\}$ 的公差为 $d$, 则 $a_n = 1 + (n-1)d$. $\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots$ (1 分)

所以 $a_{n+1}a_n = (1+nd)[1+(n-1)d] = d^2n^2 + (2-d)dn + 1-d = 4n^2 + \lambda$, $\cdots\cdots\cdots\cdots\cdots$ (4 分)

所以 $\begin{cases} d^2=4, \\ (2-d)d=0, \\ 1-d=\lambda, \end{cases}$ 解得 $\begin{cases} d=2, \\ \lambda=-1. \end{cases}$ $\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots$ (6 分)

所以 $\{a_n\}$ 的通项公式为 $a_n = 1 + 2(n-1) = 2n-1$. $\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots$ (7 分)

(2) 由 (1) 可得 $a_{n+1}a_n = 4n^2 - 1$. $\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots$ (8 分)

所以 $\frac{1}{a_{n+1}a_n} = \frac{1}{4n^2-1} = \frac{1}{(2n-1)(2n+1)} = \frac{1}{2}\left(\frac{1}{2n-1} - \frac{1}{2n+1}\right)$, $\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots$ (11 分)

则 $S_n = \frac{1}{2}\left(\frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \cdots + \frac{1}{2n-1} - \frac{1}{2n+1}\right)$

$= \frac{1}{2}\left(1 - \frac{1}{2n+1}\right)$

$= \frac{n}{2n+1}$. $\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots$ (15 分)