《2025年练习生高中数学选择性必修第二册人教版》

34. [2022·浙江嘉兴第五高级中学高二期中]已知数列$\{ a_n \}$中,$a_1 = 2$,且满足$a_n = \frac {n}{n - 1}a_{n - 1} + n$($n\in N^*$,$n \geqslant 2$).
(1) 求$a_2$,$a_3$的值;
(2) 证明:数列$\{ \frac {a_n}{n} \}$是等差数列,并求数列$\{ a_n \}$的通项公式;
(3) 若$a_n ＜ \lambda × 2^n$恒成立,求实数$\lambda$的取值范围.

答案： 34.【解】
(1)由题意得$a_{2} = \frac {2}{2 - 1} × 2 + 2 = 6$,$a_{3} = \frac {3}{3 - 1} × 6 + 3 = 12$.
(2)$\because \frac {a_{n}}{n} - \frac {a_{n - 1}}{n - 1} = \frac {n - 1}{n}a_{n - 1} + n - \frac {n - 1}{n - 1}a_{n - 1} = 1(n \geqslant 2)$
(利用等差数列的定义证明),
$\therefore$数列$\{ \frac {a_{n}}{n}\}$是首项为2,公差为1的等差数列,
$\therefore \frac {a_{n}}{n} = n + 1$,$\therefore a_{n} = n^{2} + n$.
(3)由
(2)可知$a_{n} = n^{2} + n$. $\because n^{2} + n ＜ \lambda × 2^{n}$,$\therefore \frac {n^{2} + n}{2^{n}} ＜ \lambda$.
令$f(n) = \frac {n^{2} + n}{2^{n}}(n \in N^{*})$,则$f(n)_{\max} ＜ \lambda$.
$\because f(n) - f(n - 1) = \frac {n^{2} + n}{2^{n}} - \frac {(n - 1)^{2} + n - 1}{2^{n - 1}} = \frac {3n - n^{2}}{2^{n}}$,
$\therefore$当$n \leqslant 3$时,$f(n) \geqslant f(n - 1)$;
当$n ＞ 3$时,$f(n) ＜ f(n - 1)$.
$\therefore$当$n = 2$或$n = 3$时,$f(n)$有最大值$\frac {3}{2}$,$\therefore \lambda ＞ \frac {3}{2}$.

35. [2022·河南部分学校高三联考]已知数列$\{ a_n \}$满足$a_1 = 3$,$a_2 = 2$,且$\{ \frac {1}{a_n - 1} \}$为等差数列.
(1) 求$\{ a_n \}$的通项公式;
(2) 求满足不等式$a_1 a_2 ·s a_n ＜ 2 023$的最大正整数$n$.

答案： 35.【解】
(1)由题意,得$\frac {1}{a_{1} - 1} = \frac {1}{3 - 1} = \frac {1}{2}$,$\frac {1}{a_{2} - 1} = \frac {1}{2 - 1} = 1$.
因为$\{ \frac {1}{a_{n} - 1}\}$为等差数列,
所以公差$d = \frac {1}{a_{2} - 1} - \frac {1}{a_{1} - 1} = 1 - \frac {1}{2} = \frac {1}{2}$,
所以$\frac {1}{a_{n} - 1} = \frac {1}{2} + \frac {1}{2}(n - 1) = \frac {n}{2}$,故$a_{n} = \frac {n + 2}{n}$.
(2)由
(1)得,$a_{1}a_{2}·s a_{n} = \frac {3}{1} × \frac {4}{2} × \frac {5}{3} × ·s × \frac {n + 2}{n} = \frac {(n + 1)(n + 2)}{2}$,
所以$\frac {(n + 1)(n + 2)}{2} ＜ 2023$,即$(n + 1)(n + 2) ＜ 4046$.
因为$63 × 64 = 4032 ＜ 4046$,$64 × 65 = 4160 ＞ 4046$,
所以满足不等式$a_{1}a_{2}·s a_{n} ＜ 2023$的最大正整数$n$为62.

36. 已知数列$\{ a_n \}$的前$n$项和$S_n = n^2 + 3n + 2$,判断$\{ a_n \}$是否为等差数列.

答案： 36.【解】当$n = 1$时,$a_{1} = S_{1} = 6$;
当$n \geqslant 2$时,$a_{n} = S_{n} - S_{n - 1} = (n^{2} + 3n + 2) - [(n - 1)^{2} + 3(n - 1) + 2] = 2n + 2$.
$\because$当$n = 1$时,$a_{1} = 6$不满足上式,
$\therefore a_{n} = \begin{cases}6, n = 1, \\2n + 2, n \geqslant 2.\end{cases}$
$\because a_{2} - a_{1} \neq a_{3} - a_{2}$,
$\therefore \{ a_{n}\}$不是等差数列.
易错规避 本题容易产生如下错【解析】$\because a_{n} = S_{n} - S_{n - 1} = (n^{2} + 3n + 2) - [(n - 1)^{2} + 3(n - 1) + 2] = 2n + 2$,$a_{n + 1} - a_{n} = [2(n + 1) + 2] - (2n + 2) = 2$,$\therefore$数列$\{ a_{n}\}$是等差数列. 这是因为忽视了$a_{n} = 2n + 2$中$n$的最小值是2,因此使用$a_{n + 1} - a_{n}$时$n$的最小值是2,只能得到$a_{3} - a_{2} = a_{4} - a_{3} = ·s$,而不含$a_{2} - a_{1}$.