2025學(xué)年江蘇省徐州市云龍區(qū)高二上學(xué)期期末數(shù)學(xué)考試（數(shù)列與微分方程綜合訓(xùn)練）

2025學(xué)年江蘇省徐州市云龍區(qū)高二上學(xué)期期末數(shù)學(xué)考試（數(shù)列與微分方程綜合訓(xùn)練）一、選擇題（本大題共10小題，每小題5分，共50分）1.已知數(shù)列$\{a_{n}\}$滿足$a_{1}=1$，且對(duì)于任意$n\geq2$，有$a_{n}=2a_{n-1}-1$，則數(shù)列$\{a_{n}\}$的通項(xiàng)公式為（）A.$a_{n}=2^{n}-1$B.$a_{n}=2^{n+1}-1$C.$a_{n}=2^{n}-2$D.$a_{n}=2^{n}-3$2.設(shè)數(shù)列$\{a_{n}\}$是等差數(shù)列，且$a_{1}=3$，$a_{6}=15$，則數(shù)列$\{a_{n}\}$的前10項(xiàng)和$S_{10}$為（）A.100B.150C.200D.2503.若數(shù)列$\{a_{n}\}$是等比數(shù)列，且$a_{1}=2$，$a_{3}=16$，則數(shù)列$\{a_{n}\}$的公比為（）A.2B.4C.8D.164.已知數(shù)列$\{a_{n}\}$滿足$a_{1}=1$，且對(duì)于任意$n\geq2$，有$a_{n}=3a_{n-1}$，則數(shù)列$\{a_{n}\}$的通項(xiàng)公式為（）A.$a_{n}=3^{n-1}$B.$a_{n}=3^{n}$C.$a_{n}=3^{n-2}$D.$a_{n}=3^{n+1}$5.若數(shù)列$\{a_{n}\}$是等差數(shù)列，且$a_{1}=2$，$d=3$，則數(shù)列$\{a_{n}\}$的第10項(xiàng)$a_{10}$為（）A.25B.27C.30D.336.設(shè)數(shù)列$\{a_{n}\}$是等比數(shù)列，且$a_{1}=1$，$a_{4}=81$，則數(shù)列$\{a_{n}\}$的公比為（）A.3B.9C.27D.817.已知數(shù)列$\{a_{n}\}$滿足$a_{1}=1$，且對(duì)于任意$n\geq2$，有$a_{n}=2a_{n-1}+1$，則數(shù)列$\{a_{n}\}$的通項(xiàng)公式為（）A.$a_{n}=2^{n}-1$B.$a_{n}=2^{n+1}-1$C.$a_{n}=2^{n}-2$D.$a_{n}=2^{n}-3$8.若數(shù)列$\{a_{n}\}$是等差數(shù)列，且$a_{1}=3$，$a_{8}=39$，則數(shù)列$\{a_{n}\}$的公差$d$為（）A.3B.4C.5D.69.設(shè)數(shù)列$\{a_{n}\}$是等比數(shù)列，且$a_{1}=2$，$a_{6}=64$，則數(shù)列$\{a_{n}\}$的公比$q$為（）A.2B.4C.8D.1610.已知數(shù)列$\{a_{n}\}$滿足$a_{1}=1$，且對(duì)于任意$n\geq2$，有$a_{n}=3a_{n-1}+2$，則數(shù)列$\{a_{n}\}$的通項(xiàng)公式為（）A.$a_{n}=3^{n}-1$B.$a_{n}=3^{n+1}-1$C.$a_{n}=3^{n}-2$D.$a_{n}=3^{n+1}-2$二、填空題（本大題共10小題，每小題5分，共50分）1.若數(shù)列$\{a_{n}\}$是等差數(shù)列，且$a_{1}=2$，$d=3$，則$a_{10}=______$。2.若數(shù)列$\{a_{n}\}$是等比數(shù)列，且$a_{1}=4$，$q=2$，則$a_{5}=______$。3.若數(shù)列$\{a_{n}\}$是等差數(shù)列，且$a_{1}=1$，$a_{4}=11$，則公差$d=______$。4.若數(shù)列$\{a_{n}\}$是等比數(shù)列，且$a_{1}=3$，$a_{3}=27$，則公比$q=______$。5.數(shù)列$\{a_{n}\}$的前5項(xiàng)和為$S_{5}=60$，且$a_{1}=3$，則公差$d=______$。6.數(shù)列$\{a_{n}\}$的前6項(xiàng)和為$S_{6}=252$，且$a_{1}=2$，則公比$q=______$。7.若數(shù)列$\{a_{n}\}$是等差數(shù)列，且$a_{1}=1$，$a_{6}=21$，則前6項(xiàng)和$S_{6}=______$。8.若數(shù)列$\{a_{n}\}$是等比數(shù)列，且$a_{1}=2$，$a_{4}=16$，則前4項(xiàng)和$S_{4}=______$。9.若數(shù)列$\{a_{n}\}$滿足$a_{1}=1$，$a_{2}=3$，且對(duì)于任意$n\geq3$，有$a_{n}=2a_{n-1}-a_{n-2}$，則$a_{3}=______$。10.若數(shù)列$\{a_{n}\}$滿足$a_{1}=2$，$a_{2}=5$，且對(duì)于任意$n\geq3$，有$a_{n}=3a_{n-1}-2a_{n-2}$，則$a_{3}=______$。三、解答題（本大題共3小題，共40分）1.（本題10分）已知數(shù)列$\{a_{n}\}$是等差數(shù)列，且$a_{1}=2$，$d=3$，求證：$a_{n}+a_{n+1}=a_{1}+a_{n+2}$。2.（本題10分）已知數(shù)列$\{a_{n}\}$是等比數(shù)列，且$a_{1}=3$，$q=2$，求證：$a_{n}+a_{n+1}=a_{1}q^{n+1}$。3.（本題20分）已知數(shù)列$\{a_{n}\}$滿足$a_{1}=1$，$a_{2}=2$，且對(duì)于任意$n\geq3$，有$a_{n}=3a_{n-1}-2a_{n-2}$，求證：數(shù)列$\{a_{n}\}$是等差數(shù)列，并求出公差$d$。四、解答題（本大題共3小題，共40分）4.（本題10分）已知數(shù)列$\{a_{n}\}$是等差數(shù)列，且$a_{1}=5$，$a_{6}=23$，求該數(shù)列的通項(xiàng)公式和前10項(xiàng)和$S_{10}$。5.（本題10分）已知數(shù)列$\{a_{n}\}$是等比數(shù)列，且$a_{1}=4$，$a_{4}=64$，求該數(shù)列的公比$q$和前5項(xiàng)和$S_{5}$。6.（本題20分）已知數(shù)列$\{a_{n}\}$滿足$a_{1}=2$，$a_{2}=3$，且對(duì)于任意$n\geq3$，有$a_{n}=2a_{n-1}+a_{n-2}$，求證：數(shù)列$\{a_{n}\}$是等比數(shù)列，并求出公比$q$。本次試卷答案如下：一、選擇題1.A解析：由遞推關(guān)系$a_{n}=2a_{n-1}-1$，可得$a_{2}=2a_{1}-1=1$，$a_{3}=2a_{2}-1=1$，以此類推，可得$a_{n}=2^{n}-1$。2.D解析：由等差數(shù)列的前$n$項(xiàng)和公式$S_{n}=\frac{n(a_{1}+a_{n})}{2}$，代入$a_{1}=3$，$a_{6}=15$，得$S_{10}=\frac{10(3+15)}{2}=250$。3.B解析：由等比數(shù)列的通項(xiàng)公式$a_{n}=a_{1}q^{n-1}$，代入$a_{1}=2$，$a_{3}=16$，得$q=2$。4.A解析：由遞推關(guān)系$a_{n}=3a_{n-1}$，可得$a_{2}=3a_{1}=3$，$a_{3}=3a_{2}=9$，以此類推，可得$a_{n}=3^{n-1}$。5.C解析：由等差數(shù)列的通項(xiàng)公式$a_{n}=a_{1}+(n-1)d$，代入$a_{1}=2$，$d=3$，得$a_{10}=2+(10-1)\times3=30$。6.A解析：由等比數(shù)列的通項(xiàng)公式$a_{n}=a_{1}q^{n-1}$，代入$a_{1}=1$，$a_{4}=81$，得$q=3$。7.A解析：由遞推關(guān)系$a_{n}=2a_{n-1}+1$，可得$a_{2}=2a_{1}+1=3$，$a_{3}=2a_{2}+1=7$，以此類推，可得$a_{n}=2^{n}-1$。8.B解析：由等差數(shù)列的通項(xiàng)公式$a_{n}=a_{1}+(n-1)d$，代入$a_{1}=3$，$a_{8}=39$，得$d=4$。9.B解析：由等比數(shù)列的通項(xiàng)公式$a_{n}=a_{1}q^{n-1}$，代入$a_{1}=2$，$a_{6}=64$，得$q=4$。10.A解析：由遞推關(guān)系$a_{n}=3a_{n-1}+2$，可得$a_{2}=3a_{1}+2=5$，$a_{3}=3a_{2}+2=17$，以此類推，可得$a_{n}=3^{n}-1$。二、填空題1.32解析：由等差數(shù)列的通項(xiàng)公式$a_{n}=a_{1}+(n-1)d$，代入$a_{1}=2$，$d=3$，得$a_{10}=2+(10-1)\times3=32$。2.32解析：由等比數(shù)列的通項(xiàng)公式$a_{n}=a_{1}q^{n-1}$，代入$a_{1}=4$，$q=2$，得$a_{5}=4\times2^{4}=32$。3.3解析：由等差數(shù)列的通項(xiàng)公式$a_{n}=a_{1}+(n-1)d$，代入$a_{1}=1$，$a_{4}=11$，得$d=3$。4.3解析：由等比數(shù)列的通項(xiàng)公式$a_{n}=a_{1}q^{n-1}$，代入$a_{1}=3$，$a_{3}=27$，得$q=3$。5.5解析：由等差數(shù)列的前$n$項(xiàng)和公式$S_{n}=\frac{n(a_{1}+a_{n})}{2}$，代入$a_{1}=3$，$S_{5}=60$，得$a_{5}=a_{1}+(5-1)d=3+4d=15$，解得$d=5$。6.4解析：由等比數(shù)列的前$n$項(xiàng)和公式$S_{n}=\frac{a_{1}(q^{n}-1)}{q-1}$，代入$a_{1}=2$，$S_{6}=252$，得$q=4$。7.42解析：由等差數(shù)列的通項(xiàng)公式$a_{n}=a_{1}+(n-1)d$，代入$a_{1}=1$，$a_{6}=21$，得$d=4$，則前6項(xiàng)和$S_{6}=6\times\frac{1+21}{2}=42$。8.30解析：由等比數(shù)列的前$n$項(xiàng)和公式$S_{n}=\frac{a_{1}(q^{n}-1)}{q-1}$，代入$a_{1}=2$，$a_{4}=16$，得$q=4$，則前4項(xiàng)和$S_{4}=4\times\frac{2(4^{3}-1)}{4-1}=30$。9.7解析：由遞推關(guān)系$a_{n}=2a_{n-1}-a_{n-2}$，代入$a_{1}=1$，$a_{2}=3$，得$a_{3}=2a_{2}-a_{1}=5$。10.17解析：由遞推關(guān)系$a_{n}=3a_{n-1}-2a_{n-2}$，代入$a_{1}=2$，$a_{2}=5$，得$a_{3}=3a_{2}-2a_{1}=7$。三、解答題1.（本題10分）證明：已知數(shù)列$\{a_{n}\}$是等差數(shù)列，且$a_{1}=2$，$d=3$，則$a_{n}=2+(n-1)\times3$，$a_{n+1}=2+n\times3$，$a_{n+2}=2+(n+1)\times3$。所以$a_{n}+a_{n+1}=2+(n-1)\times3+2+n\times3=2+2n+3=2+2(n+1)+1=2+a_{1}+a_{n+2}$。2.（本題10分）證明：已知數(shù)列$\{a_{n}\}$是等比數(shù)列，且$a_{1}=3$，$q=2$，則$a_{n}=3\times2^{n-1}$，$a_{n+1}=3\times2^{n}$。所以$a_{n}+a_{n+1}=3\times2^{n-1}+3\times2^{n}=3\times2^{n-1}(1+2)=3\times2^{n-1}\times3=3\times3^{n-1}=a_{1}q^{n+1}$。3.（本題20分）證明：已知數(shù)列$\{a_{n}\}$滿足$a_{1}=1$，$a_{2}=2$，且對(duì)于任意$n\geq3$，有$a_{n}=3a_{n-1}-2a_{n-2}$。所以$a_{3}=3a_{2}-2a_{1}=3\times2-2\times1=4$，$a_{4}=3a_{3}-2a_{2}=3\times4-2\times2=8$，以此類推，可得$a_{n}=3^{n}-2$。由等差數(shù)列的通項(xiàng)公式$a_{n}=a_{1}+(n-1)d$，代入$a_{1}=1$，$a_{n}=3^{n}-2$，得$d=2$。四、解答題4.（本題10分）解：已知數(shù)列$\{a_{n}\}$是等差數(shù)列，且$a_{1}=5$，$a_{6}=23$，則公差$d=\frac{a_{6}-a_{1}}{6-1}=3$。所以通項(xiàng)公式為$a_{n}=5+(n-1)\times3=3n+2$，$S_{10}=\frac{10(a_{1}+a_{10})}{2}=10\times\frac{5+32}{2}=175$。5.（本題10分）解：已知數(shù)列$\{a_{n}\}$是等比數(shù)列，且$a_{1}=4$，$a_{4}=64$，則公比$q=\sqrt[3]{\frac{a_{4}}{a_{1}}}=\sqrt[3]{\frac{64}{4}}=4$。所以公比$q=4$，$S_{5}=\frac{a_{1}(q^{5}-1)}{q-1}=\frac