《2025年学霸高考黑题数学人教版》

变式训练 1.(2025·山西忻州模拟)已知数列$\{ a_{n}\} $的前n项和$S_{n}$满足$S_{n}=2^{n+1}+1.$
(1)求$\{ a_{n}\} $的通项公式;
(2)若$\forall n∈N^{*},a_{2n}＞\lambda n· 3^{n}$恒成立,求实数λ的取值范围.

答案：
(1)解：$S_{n} = 2^{n + 1} + 1$，则当$n \geqslant 2$时，$a_{n} = S_{n} - S_{n - 1} = 2^{n + 1} + 1 - (2^{n} + 1) = 2^{n}$，当$n = 1$时，$a_{1} = S_{1} = 2^{2} + 1 = 5$，不符合$a_{n} = 2^{n}$，所以$a_{n} = \begin{cases} 5, n = 1, \\ 2^{n}, n \geqslant 2. \end{cases}$
(2)解：因为$\forall n \in N^{*}$，$a_{2n} ＞ \lambda n · 3^{n}$恒成立，所以$\forall n \in N^{*}$，$\lambda ＜ \frac{a_{2n}}{n · 3^{n}} = \frac{1}{n} · (\frac{4}{3})^{n}$.令$c_{n} = \frac{1}{n} · (\frac{4}{3})^{n}$，则$c_{n} = \frac{4}{3}$，当$n \geqslant 2$时，不妨设$\{ c_{n}\}$的第$n$项最小，只需令$\begin{cases} \frac{1}{n} · (\frac{4}{3})^{n} \leqslant \frac{1}{n + 1} · (\frac{4}{3})^{n + 1} \\ \frac{1}{n} · (\frac{4}{3})^{n} \leqslant \frac{1}{n - 1} · (\frac{4}{3})^{n - 1} \end{cases}$，解得$3 \leqslant n \leqslant 4$，
又$c_{3} = c_{4} = \frac{64}{81} · \frac{4}{3} = \frac{256}{243}$，所以$\{ c_{n}\}$的最小值为$\frac{64}{81} · \frac{4}{3} = \frac{256}{243}$，所以$\lambda ＜ \frac{64}{81}$，即$\lambda$的取值
范围是$( - \infty,\frac{64}{81})$.

变式训练 2.(2025·河北秦皇岛一模)设$S_{n}$为数列$\{ a_{n}\} $的前n项和,已知$a_{1}=4,\{ \frac {a_{n}+S_{n}}{2a_{n}}\} $是公比为 2 的等比数列.
(1)证明:$\{ a_{n}\} $是等比数列;
(2)求$\{ a_{n}\} $的通项公式以及$S_{n}$;
(3)设$b_{n}=(n-5)a_{n}$,若$\exists n∈N^{*}$,使$4^{m}+2^{m-2}≤b_{n}$,求 m 的取值范围.

答案：
(1)证明：由已知可得，数列$\{\frac{a_{n} + S_{n}}{2a_{n}}\}$是首项为$1$，公比为$2$
的等比数列，则$\frac{a_{n} + S_{n}}{2a_{n}} = 2^{n - 1}$，即$S_{n} = (2^{n} - 1)a_{n}$①，则$S_{n + 1} = (2^{n + 1} - 1)a_{n + 1}$②，②-①得$a_{n + 1} = (2^{n + 1} - 1)a_{n + 1} - (2^{n} - 1)a_{n}$，即$(2^{n + 1} - 2)a_{n + 1} = (2^{n} - 1)a_{n}$，可得$a_{n + 1} = \frac{1}{2}a_{n}$，又$a_{1} = 4 \neq 0$，$\therefore \{ a_{n}\}$是等比数列.
(2)解：由
(1)知$a_{n} = 4 · (\frac{1}{2})^{n - 1} = (\frac{1}{2})^{n - 3}$，则$S_{n} = (2^{n} - 1)a_{n} = 8 - \frac{1}{2^{n - 3}}$.
(3)解：由$b_{n} = (n - 5)a_{n}$，且$a_{n} = (\frac{1}{2})^{n - 3}$，得$b_{n} = \frac{n - 5}{2^{n - 3}}$，当$n \leqslant 5$时，
$b_{n} \leqslant 0$，当$n \geqslant 6$时，$b_{n} ＞ 0$，$\therefore \frac{b_{n + 1}}{b_{n}} = \frac{\frac{n - 4}{2^{n - 2}}}{\frac{n - 5}{2^{n - 3}}} = \frac{1}{2} · \frac{n - 4}{n - 5} = \frac{1}{2}(1 + \frac{1}{n - 5})$，若
$n = 6$，则$b_{7} = b_{6} = \frac{1}{8}$，若$n \geqslant 7$，则$\frac{b_{n + 1}}{b_{n}} = \frac{1}{2}(1 + \frac{1}{n - 5}) \leqslant \frac{3}{4} ＜ 1$，可得$b_{n + 1} ＜ b_{n}$，因此数列$\{ b_{n}\}$的最大项为$b_{7} = b_{6} = \frac{1}{8}$，由$\exists n \in N^{*}$，使
$4^{m} + 2^{m - 2} \leqslant b_{n}$，得$4^{m} + 2^{m - 2} \leqslant \frac{1}{8}$，即$8 · (2^{m})^{2} + 2 · 2^{m} - 1 \leqslant 0$，整理得
$(4 · 2^{m} - 1)(2 · 2^{m} + 1) \leqslant 0$，则$2^{m} \leqslant \frac{1}{4}$，即$m \leqslant - 2$，$\therefore m$的取值范围是
$( - \infty, - 2]$.

变式训练 3.(2025·河南洛阳模拟)已知数列$\{ a_{n}\} $的首项为 1,其前n项和为$S_{n}$,等比数列$\{ b_{n}\} $是首项为 1 的递增数列,若$3na_{n+1}-6S_{n}=n(n+1)(n+2),8b_{2}+2b_{4}=b_{6}.$
(1)求证:数列$\{ \frac {a_{n}}{n}\} $是等差数列;
(2)求数列$\{ (-1)^{n}· 2a_{n}\} $的前n项和;
(3)求使得$a_{n}≥b_{n}$成立的最大整数n.

答案：
(1)证明：$3na_{n + 1} - 6S_{n} = n(n + 1)(n + 2)$①，$\therefore$当$n \geqslant 2$
时，$3(n - 1)a_{n} - 6S_{n - 1} = (n - 1)n(n + 1)$②，式子①-②，化简得$na_{n + 1} -(n + 1)a_{n} = n(n + 1)$，两边同时除以$n(n + 1)$得$\frac{a_{n + 1}}{n + 1} - \frac{a_{n}}{n} = 1(n \geqslant 2)$.
又$3na_{n + 1} - 6S_{n} = n(n + 1)(n + 2)$中，令$n = 1$得$3a_{2} - 6S_{1} = 6$，即$3a_{2} - 6a_{1} = 6$.
又$a_{1} = 1$，故$a_{2} = 4$，$\therefore \frac{a_{2}}{2} - \frac{a_{1}}{1} = 1$，故对$\forall n \in N^{*}$，$\frac{a_{n + 1}}{n + 1} - \frac{a_{n}}{n} = 1$，$\therefore$数列$\{\frac{a_{n}}{n}\}$
是首项为$1$，公差为$1$的等差数列.
(2)解：由
(1)得$\frac{a_{n}}{n} = 1 + (n - 1) × 1 = n$，则$a_{n} = n^{2}$，设数列
$\{(- 1)^{n} · 2a_{n}\}$的前$n$项和为$T_{n}$，当$n$为偶数时，$T_{n} = 2[ - 1^{2} + 2^{2} - 3^{2} +·s + ( - 1)^{n} · n^{2}]$，$\therefore n^{2} - (n - 1)^{2} = 2n - 1$，$\therefore T_{n} = 2[3 + 7 + ·s + (2n -1)] = 2 · \frac{(3 + 2n - 1) · \frac{n}{2}}{2} = n(n + 1) = n^{2} + n$，当$n$为奇数时，$n + 1$为偶数，$n + 1$为偶数.
$T_{n} = T_{n + 1} - 2 · (- 1)^{n + 1}(n + 1)^{2} = (n + 1)^{2} + n + 1 - 2(n + 1)^{2} = - n^{2} - n$，
$\therefore T_{n} = \begin{cases} n^{2} + n, n = 2k, k \in N^{*}, \\ - n^{2} - n, n = 2k - 1, k \in N^{*}. \end{cases}$
(3)解：设等比数列$\{ b_{n}\}$的公比为$q(q \neq 0)$，$\because b_{1} = 1$，$\therefore$由$8b_{2} + 2b_{4} =b_{6} \Rightarrow 8q + 2q^{3} = q^{5} \Rightarrow q = 0$或$q = \pm 2$.又$\because$数列$\{ b_{n}\}$是递增数列，$\therefore q = 2 \Rightarrowb_{n} = 2^{n - 1}$.由
(2)知，$a_{n} \geqslant b_{n}$，即$n^{2} \geqslant 2^{n - 1}$，令$c_{n} = n^{2} - 2^{n - 1}$，则$d_{n} =c_{n + 1} - c_{n} = (n + 1)^{2} - 2^{n} + 2^{n - 1} = 2n + 1 - 2^{n - 1}$，$d_{n + 1} - d_{n} = 2n + 3 - 2^{n}$，$\therefore$当$n = 1$时，$d_{2} - d_{1} = 2 - 1 = 1 ＞ 0$，当$n = 2$时，$d_{2} = d_{3}$，当$n \geqslant 3$时，
$d_{n + 1} ＜ d_{n}$，即有$d_{1} ＜ d_{2} = d_{3} ＞ d_{4} ＞ d_{5} ＞ ·s$.又$d_{1} = 2 ＞ 0$，$d_{2} = d_{3} = 3 ＞ 0$，$d_{4} =1 ＞ 0$，$d_{5} = - 5 ＜ 0$，$\therefore$当$n \geqslant 5$时，$d_{n} ＜ 0$，$\therefore c_{1} ＜ c_{2} ＜ c_{3} ＜ c_{4} ＜ c_{5}$，$c_{5} ＞ c_{6} ＞ c_{7} ＞ ·s$
又$c_{1} = 0$，$c_{6} = 6^{2} - 2^{5} ＞ 0$，$c_{7} = 7^{2} - 2^{6} ＜ 0$，$\therefore a_{6} ＞ b_{6}$，当$n \geqslant 7$时，$a_{n} ＜ b_{n}$，故使
得$a_{n} \geqslant b_{n}$成立的最大整数$n$为$6$.