《2025年学霸高考黑题数学人教版》

典型例题5. (2022·天津)设$\{a_n\}$是等差数列，$\{b_n\}$是等比数列，且$a_1 = b_1 = a_2 - b_2 = a_3 - b_3 = 1$.
(1)求$\{a_n\}$与$\{b_n\}$的通项公式；
(2)设$\{a_n\}$的前$n$项和为$S_n$，求证：$(S_{n + 1}+a_{n + 1})b_n = S_{n + 1}b_{n + 1}-S_nb_n$；
(3)求$\sum_{k = 1}^{2n}[a_{k + 1}-(-1)^ka_k]b_k$.

答案： 典型例题5.
(1) 解：设$\{ a_n \}$的公差为$d$，$\{ b_n \}$的公比为$q$，则$a_n = 1 + (n - 1)d$，$b_n = q^{n - 1}$，由$a_2 - b_2 = a_3 - b_3 = 1$可得$\begin{cases} 1 + d - q = 1, \\ 1 + 2d - q^2 = 1 \end{cases} \Rightarrow \begin{cases} d = q, \\ 2d = q^2 \end{cases} \Rightarrow d = q = 2(d = q = 0$舍去$)$，所以$a_n = 2n - 1$，$b_n = 2^{n - 1}(n \in \mathbf{N}^*)$.
(2) 证明：因为$b_{n + 1} = 2b_n \neq 0$，所以要证$(S_{n + 1} + a_{n + 1})b_n = S_{n + 1} · 2b_n - S_n b_n$，即证$(S_{n + 1} + a_{n + 1})b_n = 2S_{n + 1}b_n - S_n b_n$，即证$a_{n + 1} = S_{n + 1} - S_n$，而$a_{n + 1} = S_{n + 1} - S_n$显然成立，所以$(S_{n + 1} + a_{n + 1})b_n = S_{n + 1} · b_{n + 1} - S_n b_n$.
(3) 解：因为$[a_{2k} - (-1)^{2k - 1}a_{2k - 1}]b_{2k - 1} + [a_{2k + 1} - (-1)^{2k}a_{2k}]b_{2k} = (4k - 1 + 4k - 3) × 2 × 4^{k - 1} + [4k + 1 - 4(k - 1)] × 2^{2k - 1} = 2k · 4^k$，所以$\sum_{k = 1}^{n} [a_{k + 1} - (-1)^k a_k] · b_k = \sum_{k = 1}^{n} [a_{2k} - (-1)^{2k - 1}a_{2k - 1}]b_{2k - 1} + [a_{2k + 1} - (-1)^{2k}a_{2k}]b_{2k} = \sum_{k = 1}^{n} 2k · 4^k$. 设$T_n = \sum_{k = 1}^{n} 2k · 4^k = 2 × 4 + 4 × 4^2 + 6 × 4^3 + ·s + 2n × 4^n$，则$4T_n = 2 × 4^2 + 4 × 4^3 + 6 × 4^4 + ·s + 2n × 4^{n + 1}$，作差得$-3T_n = 2 × (4 + 4^2 + 4^3 + 4^4 + ·s + 4^n) - 2n × 4^{n + 1} = 2 × \frac{4 × (1 - 4^n)}{1 - 4} - 2n × 4^{n + 1} = \frac{2 × (4 - 4^{n + 1})}{3} - 2n × 4^{n + 1} = \frac{(6 - 2)4^{n + 1} - 8}{3}$，所以$T_n = \frac{(6 - 2)4^{n + 1} + 8}{9} = \frac{(6n - 2)4^{n + 1} + 8}{9}$，所以$\sum_{k = 1}^{n} [a_{k + 1} - (-1)^k a_k]b_k = \frac{(6n - 2)4^{n + 1} + 8}{9}$.

变式训练7. (2025·福建三明三模)若数列$\{a_n\}$满足$a_{n + 1}+a_n = 2n + 1(n\in\mathbf{N}^*)$，$a_4 = 2$，则$a_1 + a_2 + a_3+·s+a_{17}=$ (A)

A.$155$
B.$156$
C.$203$
D.$204$

答案： 变式训练7. A 解析：由$a_{n + 1} + a_n = 2n + 1(n \in \mathbf{N}^*)$，则$a_{n + 1} - a_n = 2$，故奇数项成等差数列，偶数项成等差数列，由$a_4 = 2$，则$a_3 + a_4 = 7$，$a_3 = 5$，则$a_{17} = a_3 + \frac{17 - 3}{2} × 2 = 19$，故$a_1 + a_2 + a_3 + ·s + a_{17} = (a_1 + a_2) + (a_3 + a_4) + ·s + (a_{15} + a_{16}) + a_{17} = 3 + 7 + ·s + 2 × 15 + 1 + 19 = 3 + 7 + ·s + 31 + 19 = 155$. 故选A.

变式训练8. (2025·广东阳江模拟)已知等差数列$\{a_n\}$的首项为$1$，且$a_3$，$a_5 + 1$，$2a_6$成等比数列，则数列$\{(-1)^na_n\}$的前$2025$项和为 (A)

A.$-1013$
B.$-505$
C.$505$
D.$1013$

答案： 变式训练8. A 解析：设公差为$d$. 因为$a_3$，$a_5 + 1$，$a_6$成等比数列，所以$(a_5 + 1)^2 = a_3 × 2a_6$，则$(1 + 4d + 1)^2 = (1 + 2d) × 2(1 + 5d)$，解得$d = 1$或$d = -\frac{1}{2}$，当$d = -\frac{1}{2}$时，$a_3 = 1 + 2 × (-\frac{1}{2}) = 0$，此时与$a_3$，$a_5 + 1$，$a_6$成等比数列矛盾，故排除，当$d = 1$时，$a_n = 1 + n - 1 = n$，此时令$b_n = (-1)^n a_n = (-1)^n n$，而其前$2025$项和为$-1 + 2 - 3 + 4 - ·s - 2025 = (-1 + 2) + (-3 + 4) + ·s - 2025 = 1012 × 1 - 2025 = -1013$.

变式训练9. (2025·江苏南京师大附中月考)设数列$\{a_n\}$满足$a_1 = 1$，$a_2 = 2$，$a_{n + 2}=\begin{cases}a_n + 1,n为奇数,\\2a_n,n为偶数\end{cases}(n\in\mathbf{N}^*)$，令$b_n = (\log_2a_{2n})^2·\sin(a_{2n - 1}·\frac{\pi}{2})$，则数列$\{b_n\}$的前$100$项和为 (B)

A.$4950$
B.$-5000$
C.$-5050$
D.$-5250$

答案： 变式训练9. B 解析：数列$\{ a_n \}$满足$a_1 = 1$，$a_2 = 2$，$a_{n + 2} = \begin{cases} a_{n + 1} + 1, n 为奇数, \\ 2a_n, n 为偶数, \end{cases}$所以数列$\{ a_{2n - 1} \}$是以$a_1 = 1$为首项，$1$为公差的等差数列，即$a_{2n - 1} = n$，数列$\{ a_{2n} \}$是以$2$为首项，$2$为公比的等比数列，即$a_{2n} = 2^n$，因此$b_n = (\log_2 2^n)^2 · \sin \frac{n\pi}{2} = n^2 \sin \frac{n\pi}{2}$，显然$\{ \sin \frac{n\pi}{2} \}$的周期为$4$，则$b_{4k - 3} + b_{4k - 2} + b_{4k - 1} + b_{4k} = (4k - 3)^2 \sin \frac{(4k - 3)\pi}{2} + (4k - 2)^2 \sin \frac{(4k - 2)\pi}{2} + (4k - 1)^2 \sin \frac{(4k - 1)\pi}{2} + (4k)^2 \sin \frac{4k\pi}{2} = (4k - 3)^2 - (4k - 2)^2 = -8(2k - 1)(k \in \mathbf{N}^*)$，令$c_n = b_{4n - 3} + b_{4n - 2} + b_{4n - 1} + b_{4n}$，则$c_n = -8(2n - 1)$，因为$c_{n + 1} - c_n = -16$，所以数列$\{ c_n \}$是等差数列，所以数列$\{ b_n \}$的前$100$项和，即数列$\{ c_n \}$的前$25$项和为$\frac{25 × [(-8) + 8 × (1 - 2 × 25)]}{2} = -5000$.

变式训练10. (2025·湖北武汉月考)佩尔数列是一个在计算机科学中有着重要应用的数列.若数列$\{P_n\}$满足$P_1 = 0$，$P_2 = 1$，且$P_{n + 2}=2P_{n + 1}+P_n$，则称数列$\{P_n\}$为佩尔数列.已知$\{P_n\}$为佩尔数列，若$P_n$为奇数，把$P_n$换成$(-1)^{\frac{n(n + 1)}{2}}n$，若$P_n$为偶数，把$P_n$换成$0$，得到数列$\{a_n\}$，若$\{a_n\}$的前$k$项和为$100$，则$k$的值可能是 (B)

A.$198$
B.$201$
C.$99$
D.$100$

答案： 变式训练10. B 解析：根据数列$\{ P_n \}$满足$P_1 = 0$，$P_2 = 1$，且$P_{n + 2} = 2P_{n + 1} + P_n$，则$P_{n + 2} - P_{n + 1} = 2(P_{n + 1} - P_n) + P_n$，则$P_{n + 2} - P_{n + 1}$与$P_{n + 1} - P_n$的奇偶性不同. 又$P_1 = 0$，$P_2 = 1$，可得$\{ P_n \}$的奇数项为偶数，偶数项为奇数，则$a_n = \begin{cases} 0, n = 4k - 3, \\ 2 - 4k, n = 4k - 2, \\ 0, n = 4k - 1, \\ 4k, n = 4k, \end{cases}$其中$k \in \mathbf{N}^*$. 所以数列$\{ a_n \}$的各项依次为$0, -2, 0, 4, 0, -6, 0, 8, ·s$，注意到$a_{4k - 3} + a_{4k - 2} + a_{4k - 1} + a_{4k} = 2$，设数列$\{ a_n \}$的前$n$项和为$S_n$，则$n = 4k$时，$S_n = 2k \Rightarrow S_{200} = 2 × 50 = 100$.

变式训练11. (2025·山东潍坊月考)已知数列$\{a_n\}$满足$a_{n + 1}+(-1)^na_n = 3n - 1(n\in\mathbf{N}^*)$，则$\{a_n\}$的前$60$项的和为2760.

答案： 变式训练11.2 760 解析：由$a_{n + 1} + (-1)^n a_n = 3n - 1$，故$a_{n + 1} - a_n = 2$，$a_{4 - 3} = 2(k - 1) + a_{4 - 3} = 2$，故$a_1 + a_3 = 3$，$a_5 + a_7 = 3$，$a_9 + a_{11} = 3$，$·s$从第一项开始，依次取$2$个相邻奇数项的和都等于$3$；$a_2 + a_4 = 13$，$a_6 + a_8 = 37$，$a_{10} + a_{12} = 61$，$·s$从第二项开始，依次取$2$个相邻偶数项的和构成以$13$为首项，以$24$为公差的等差数列. 故$S_{60} = 3 × 15 + 13 × 15 + \frac{15 × 14}{2} × 24 = 2760$.