《2025年学霸高考黑题数学人教版》

变式训练8. (2025·江苏镇江模拟)
设数列$\{ a_{n}\}$的前$n$项和为$S_{n}$,若$a_{n + 1}=2\sqrt{S_{n}}+1$,且$a_{1}=1$,则(C)

A.$a_{5}＜5$
B.$a_{5}＞10$
C.$S_{100}＞1000$
D.$S_{100}＜10000$

答案： 变式训练8.C解析：数列$\mid a_n\mid$中，由$a_{n+1}=2\sqrt{S_n}+1$，得$S_{n+1}-S_n=2\sqrt{S_n}+1$，整理得$S_{n+1}=(\sqrt{S_n}+1)^2$，则$\sqrt{S_{n+1}}=\sqrt{S_n}+1$，故数列$\{\sqrt{S_n}\}$是以$\sqrt{S_1}=\sqrt{a_1}=1$为首项，$1$为公差的等差数列，于是$\sqrt{S_n}=n\Rightarrow a_{n+1}=2n+1$，即$S_n=n^2,a_n=2n-1(n\geq2)$，而$a_1=1$满足上式，因此$S_n=n^2,a_n=2n-1,a_5=9,S_{100}=10000$，ABD错误，C正确.

变式训练9. (2025·河南三门峡三模)
已知数列$\{ a_{n}\}$的前$n$项和是$S_{n}$,若$S_{n}=(-1)^{n + 1}a_{n}+n(n\geqslant 2),n\in \mathbf{N}^{*}$,则$a_{2025}=$ (D)

A.$-1$
B.1
C.2
D.3

答案： 变式训练9.D解析：数列$\mid a_n\mid$的前$n$项和是$S_n$，若$S_n=(-1)^{n+1}a_n+n(n\geq2),n\in\mathbf{N}^*$，则当$n\geq2$时，$S_{n-1}=(-1)^na_{n-1}+n-1$，两式相减可得$a_n=(-1)^{n+1}a_n-(-1)^na_{n-1}+1(n\geq2)$，当$n=2027$时，$a_{2027}=-a_{2026}+1$，解得$a_{2026}=-1$，当$n=2026$时，$a_{2026}=a_{2026}-a_{2025}+1$，解得$a_{2025}=3$.

典型例题4. (2025·江西赣州一模)
已知数列$\{ a_{n}\}$的前$n$项和为$S_{n}$,且满足$3a_{n}=2S_{n}+1$,则$S_{5}=$ (D)

A.11
B.31
C.61
D.121

答案： 典型例题4.D解析：令$n=1$，得$3a_1=2S_1+1=2a_1+1$，得$a_1=1$，由$3a_n=2S_n+1$，当$n\geq2$时，$3a_{n-1}=2S_{n-1}+1$，两式相减，得$3a_n-3a_{n-1}=2(S_n-S_{n-1})=2a_n$，即$a_n=3a_{n-1}$，即$\frac {a_n}{a_{n-1}}=3$，所以数列$\mid a_n\mid$是以$a_1=1$为首项，$3$为公比的等比数列，所以$S_5=\frac {1×(1-3^5)}{1-3}=121$.

变式训练10. (2025·河北保定月考)
已知数列$\{ a_{n}\}$满足$a_{1}=4$,且$a_{n + 1}=2a_{n}-3$,则$a_{221}=$ (C)

A.$2^{210}-3$
B.$2^{211}-1$
C.$2^{210}+3$
D.$2^{211}+1$

答案： 变式训练10.C解析：因为$a_{n+1}=2a_n-3$，所以$a_{n+1}-3=2(a_n-3)$.因为$a_1-3=1$，所以数列$\mid a_n-3\mid$是首项为$1$，公比为$2$的等比数列，所以$a_n-3=2^{n-1}$，所以$a_n=2^{n-1}+3$，故$a_{211}=2^{210}+3$.

变式训练11. (多选)(2025·江苏南京二模)
已知数列$\{ a_{n}\}$中,$a_{3}=\frac {1}{8},a_{n}-a_{n + 1}=-3a_{n + 1}a_{n},n\in \mathbf{N}^{*}$,其前$n$项和为$S_{n}$,则(ABD)

A.$a_{1}=\frac {1}{14}$
B.$a_{n}=\frac {1}{17 - 3n}$
C.$a_{n}\geqslant a_{7}$
D.$S_{10}＜0$

答案： 变式训练11.ABD解析：由$a_n-a_{n+1}=-3a_na_{n+1}$，得$\frac {1}{a_{n+1}}-\frac {1}{a_n}=-3$，所以数列$\{\frac {1}{a_n}\}$是以$-3$为公差的等差数列，而$\frac {1}{a_3}-\frac {1}{a_1}+2×(-3),a_3=\frac {1}{8}$，所以$\frac {1}{a_1}=14$，得$a_1=\frac {1}{14}$，故A正确；所以$\frac {1}{a_n}=14+(-3)(n-1)=17-3n$，得$a_n=\frac {1}{17-3n}$，故B正确；令$\frac {1}{a_n}=17-3n=0$，解得$n=\frac {17}{3}$，$a_1,a_2,a_3,a_4,a_5$为正数，且依次递增；$a_6,a_7,·s,a_n$为负数，且依次递增，所以$a_n\geq a_6$，故C错误；$S_{10}=a_1+a_2+·s+a_{10}=\frac {1}{14}+\frac {1}{11}+\frac {1}{8}+\frac {1}{5}+\frac {1}{2}-1-\frac {1}{4}-\frac {1}{7}-\frac {1}{10}-\frac {1}{13}-\frac {1}{14}+\frac {1}{11}+\frac {1}{8}+\frac {1}{5}+\frac {1}{2}=0$，故D正确.

变式训练12. (2025·江苏南通海门中学期中)
已知数列$\{ a_{n}\}$的前$n$项和为$S_{n}$,其中$a_{1}=1$,且$S_{n + 1}-2S_{n}=2n + 1$,则$\frac {S_{7}}{a_{5}}=$ (C)

A.$\frac {369}{46}$
B.$\frac {361}{46}$
C.$\frac {367}{46}$
D.$\frac {365}{46}$

答案： 变式训练12.C解析：因为$S_{n+1}-2S_n=2n+1$，所以$S_{n+1}+2(n+1)+3=2(S_n+2n+3)$，所以数列$\{S_n+2n+3\}$是首项为$6$，公比为$2$的等比数列，所以$S_n+2n+3=6×2^{n-1}$，即$S_n=3×2^n-2n-3$，所以$\frac {S_7}{a_5}-\frac {S_5}{a_5}-\frac {367}{46}$.

变式训练13. (2025·广东广州期末)
已知数列$\{ a_{n}\}$满足$a_{1}=2,a_{n + 1}=3a_{n}+2^{n - 1},n\in \mathbf{N}^{*}$,则数列$\{ a_{n}\}$的通项公式为$a_n=3^n-2^{n-1}$.

答案： 变式训练13.$a_n=3^{n-2^{-1}}$解析：由$a_1=2,a_{n+1}=3a_n+2^{n-1},n\in\mathbf{N}^*$，可得$a_{n+1}+2^n=3(a_n+2^{n-1})$，所以$\{a_n+2^{n-1}\}$是以$3$为首项，$3$为公比的等比数列，所以$a_n+2^{n-1}=3^n$，则$a_n=3^n-2^{n-1}$.

变式训练14. (2025·天津耀华中学月考)
已知数列$\{ a_{n}\}$满足$a_{1}=1,a_{2}=\frac {1}{2}$,且$a_{n + 2}=\frac {a_{n + 1}^{2}}{a_{n}+a_{n + 1}}(n\in \mathbf{N}^{*})$,则数列$\{ a_{n}\}$的通项公式是$a_n=\frac {1}{n!}$.

答案： 变式训练14.$a_n=\frac {1}{n!}$解析：由$a_{n+2}=\frac {a_{n+1}^2}{a_n+a_{n+1}}(n\in\mathbf{N}^*)$，则$\frac {a_{n+2}}{a_{n+1}}=\frac {a_{n+1}}{a_n+a_{n+1}}+1$，又$a_1=1,a_2=\frac {1}{2}$，则$\frac {a_1}{a_2}-\frac {a_2}{a_1}=2$，故$\frac {a_n}{a_{n+1}}=2+(n-1)×1=n+1$，则有$\frac {a_{n-1}}{a_n}-\frac {a_n}{a_{n-1}}=n-1,·s,\frac {a_1}{a_2}=\frac {a_1}{a_1}=2$，且$n\geq3$，故$\frac {a_n}{a_{n-1}}×\frac {a_{n-1}}{a_{n-2}}×·s×\frac {a_2}{a_1}=n!$，即$a_n=\frac {1}{n!}(n\geq3)$，显然$n=1,2$时均满足，故$a_n=\frac {1}{n!}$.

变式训练15.
已知各项都为正数的数列$\{ a_{n}\}$满足$5a_{n + 2}+4a_{n + 1}-a_{n}=0$.
(1)证明:数列$\{ a_{n}+a_{n + 1}\}$为等比数列;
(2)若$a_{1}=\frac {1}{5},a_{2}=\frac {1}{25}$,求$\{ a_{n}\}$的通项公式.

答案： 变式训练15.
(1)证明：由各项都为正数的数列$\{a_n\}$满足$5a_{n+2}+4a_{n+1}-a_n=0$，得$a_{n+1}+a_{n+2}=\frac {1}{5}(a_{n+1}+a_n)$，即$\frac {a_{n+1}+a_{n+2}}{a_n+a_{n+1}}=\frac {1}{5}$，所以数列$\{a_n+a_{n+1}\}$是公比为$\frac {1}{5}$的等比数列.
(2)解：因为$a_1=\frac {1}{5},a_2=\frac {1}{25}$，所以$a_1+a_2=\frac {6}{25}$，由
(1)知数列$\{a_n+a_{n+1}\}$是首项为$\frac {6}{25}$，公比为$\frac {1}{5}$的等比数列，所以$a_n+a_{n+1}=\frac {6}{25}×(\frac {1}{5})^{n-1}=\frac {5+1}{5}×\frac {1}{5^{n-1}}$，于是$a_{n+1}-(\frac {1}{5})^{n+1}=-[a_n-(\frac {1}{5})^n]$.又因为$a_1-\frac {1}{5}=0$，所以$a_n-(\frac {1}{5})^n=0$，即$a_n=(\frac {1}{5})^n$.