江蘇高考數(shù)列題真題及答案

一、單項選擇題1.已知數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{1}=1\)，\(a_{n+1}=2a_{n}+1\)，則\(a_{3}\)的值為（）A.3B.7C.15D.31答案：B2.設等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和為\(S_{n}\)，若\(a_{3}=5\)，\(S_{3}=9\)，則\(a_{5}\)的值為（）A.7B.9C.11D.13答案：B3.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{2}=2\)，\(a_{5}=16\)，則公比\(q\)為（）A.\(-2\)B.2C.\(\pm2\)D.4答案：B4.數(shù)列\(zhòng)(\{a_{n}\}\)的通項公式\(a_{n}=n^{2}-9n\)，則該數(shù)列的最小項是（）A.第4項B.第5項C.第4項或第5項D.第6項答案：C5.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)的公差\(d\neq0\)，且\(a_{1}\)，\(a_{3}\)，\(a_{9}\)成等比數(shù)列，則\(\frac{a_{1}+a_{3}+a_{9}}{a_{2}+a_{4}+a_{10}}\)的值為（）A.\(\frac{13}{16}\)B.\(\frac{15}{16}\)C.\(\frac{11}{16}\)D.\(\frac{12}{16}\)答案：A6.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}+a_{2}=3\)，\(a_{3}+a_{4}=12\)，則\(a_{5}+a_{6}\)的值為（）A.24B.36C.48D.60答案：C7.數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}=\frac{1}{1-a_{n}}\)，\(a_{1}=2\)，則\(a_{2023}\)的值為（）A.2B.\(-1\)C.\(\frac{1}{2}\)D.1答案：A8.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和為\(S_{n}\)，若\(S_{10}=100\)，\(S_{100}=10\)，則\(S_{110}\)的值為（）A.\(-110\)B.0C.110D.220答案：A9.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{n}>0\)，且\(a_{5}a_{6}+a_{4}a_{7}=18\)，則\(\log_{3}a_{1}+\log_{3}a_{2}+\cdots+\log_{3}a_{10}\)的值為（）A.12B.10C.8D.2+\log_{3}5答案：B10.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=n^{2}-2n+1\)，則\(a_{n}\)的通項公式為（）A.\(a_{n}=2n-3\)B.\(a_{n}=\begin{cases}0,&n=1\\2n-3,&n\geq2\end{cases}\)C.\(a_{n}=2n-1\)D.\(a_{n}=\begin{cases}0,&n=1\\2n-1,&n\geq2\end{cases}\)答案：B二、多項選擇題1.下列關于等差數(shù)列的說法正確的是（）A.若\(\{a_{n}\}\)是等差數(shù)列，\(a_{m}=a\)，\(a_{n}=b\)，則\(a_{m+n}=\frac{na-mb}{n-m}\)（\(m\neqn\)）B.若\(\{a_{n}\}\)是等差數(shù)列，則數(shù)列\(zhòng)(\{2^{a_{n}}\}\)是等比數(shù)列C.等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和為\(S_{n}\)，若\(S_{10}=S_{20}\)，則\(S_{30}=0\)D.若\(\{a_{n}\}\)是等差數(shù)列，\(S_{n}\)是其前\(n\)項和，則\(S_{n}\)，\(S_{2n}-S_{n}\)，\(S_{3n}-S_{2n}\)仍成等差數(shù)列答案：ABCD2.已知等比數(shù)列\(zhòng)(\{a_{n}\}\)的公比為\(q\)，前\(n\)項和為\(S_{n}\)，則下列說法正確的是（）A.若\(q>1\)，則\(\{a_{n}\}\)是遞增數(shù)列B.若\(a_{1}>0\)，\(0<q<1\)，則\(\{a_{n}\}\)是遞減數(shù)列C.若\(S_{n}=3^{n}+m\)（\(m\)為常數(shù)），則\(m=-1\)D.若\(a_{5}\)，\(a_{7}\)是方程\(x^{2}+4x+2=0\)的兩根，則\(a_{6}=\pm\sqrt{2}\)答案：BC3.數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}=\begin{cases}2a_{n},&0\leqa_{n}<\frac{1}{2}\\2a_{n}-1,&\frac{1}{2}\leqa_{n}<1\end{cases}\)，若\(a_{1}=\frac{3}{5}\)，則（）A.\(a_{2}=\frac{1}{5}\)B.\(a_{3}=\frac{2}{5}\)C.\(a_{4}=\frac{4}{5}\)D.\(a_{5}=\frac{3}{5}\)答案：ABCD4.設等差數(shù)列\(zhòng)(\{a_{n}\}\)的公差為\(d\)，前\(n\)項和為\(S_{n}\)，且\(a_{1}>0\)，\(S_{6}=S_{10}\)，則（）A.\(d<0\)B.\(a_{8}=0\)C.\(S_{n}\)的最大值為\(S_{8}\)或\(S_{9}\)D.\(S_{18}=0\)答案：ACD5.已知數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和為\(S_{n}\)，且\(S_{n}=2a_{n}-1\)，則（）A.\(a_{1}=1\)B.\(a_{n}=2^{n-1}\)C.\(a_{n}=2n-1\)D.\(S_{n}=2^{n}-1\)答案：ABD6.設等比數(shù)列\(zhòng)(\{a_{n}\}\)的公比為\(q\)，其前\(n\)項積為\(T_{n}\)，并且滿足條件\(a_{1}>1\)，\(a_{9}a_{10}-1>0\)，\(\frac{a_{9}-1}{a_{10}-1}<0\)，則下列結論正確的是（）A.\(0<q<1\)B.\(a_{9}a_{11}-1<0\)C.\(T_{10}\)是數(shù)列\(zhòng)(\{T_{n}\}\)中的最大值D.使\(T_{n}>1\)成立的最大自然數(shù)\(n\)等于18答案：ABD7.已知數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{1}=1\)，\(a_{n+1}-a_{n}=2n\)，則（）A.\(a_{n}=n^{2}-n+1\)B.\(a_{n}=n^{2}+n-1\)C.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=\frac{n(n^{2}+1)}{3}\)D.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=\frac{n(n^{2}-1)}{3}\)答案：AC8.對于數(shù)列\(zhòng)(\{a_{n}\}\)，定義\(H_{n}=\frac{a_{1}+2a_{2}+\cdots+2^{n-1}a_{n}}{n}\)為\(\{a_{n}\}\)的“優(yōu)值”，現(xiàn)在已知某數(shù)列\(zhòng)(\{a_{n}\}\)的“優(yōu)值”\(H_{n}=2^{n}\)，則（）A.\(a_{1}=2\)B.\(a_{n}=n+1\)C.數(shù)列\(zhòng)(\{a_{n}\}\)是等差數(shù)列D.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=\frac{n(n+3)}{2}\)答案：ABCD9.已知數(shù)列\(zhòng)(\{a_{n}\}\)是等比數(shù)列，公比\(q\neq1\)，其前\(n\)項和為\(S_{n}\)，下列說法正確的有（）A.數(shù)列\(zhòng)(\{a_{n+1}-a_{n}\}\)也是等比數(shù)列B.數(shù)列\(zhòng)(\{\frac{1}{a_{n}}\}\)是等比數(shù)列C.若\(a_{1}=1\)，\(q=2\)，則\(S_{n}=2^{n}-1\)D.若\(a_{1}=1\)，\(q=2\)，則\(S_{2n}=4^{n}-1\)答案：ABC10.設等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和為\(S_{n}\)，若\(S_{5}<S_{6}\)，\(S_{6}=S_{7}>S_{8}\)，則下列結論正確的是（）A.\(d<0\)B.\(a_{7}=0\)C.\(S_{9}>S_{5}\)D.\(S_{6}\)與\(S_{7}\)均為\(S_{n}\)的最大值答案：ABD三、判斷題1.若數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=n^{2}\)，則\(\{a_{n}\}\)是等差數(shù)列。（）答案：對2.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}>0\)，\(q>1\)，則\(\{a_{n}\}\)一定是遞增數(shù)列。（）答案：對3.數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}-a_{n}=n\)，則\(\{a_{n}\}\)是等差數(shù)列。（）答案：錯4.等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和為\(S_{n}\)，若\(S_{n}=An^{2}+Bn\)（\(A\)、\(B\)為常數(shù)），則\(\{a_{n}\}\)是等差數(shù)列。（）答案：對5.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{3}=4\)，\(a_{7}=16\)，則\(a_{5}=8\)。（）答案：錯6.數(shù)列\(zhòng)(\{a_{n}\}\)的通項公式\(a_{n}=n+\frac{1}{n}\)，則\(\{a_{n}\}\)的最小項為\(a_{1}=2\)。（）答案：錯7.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}+a_{5}=10\)，\(a_{4}=7\)，則\(\{a_{n}\}\)的公差\(d=2\)。（）答案：對8.等比數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和為\(S_{n}\)，若\(S_{n}=3^{n}+m\)，則\(m=-1\)。（）答案：對9.若數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}=a_{n}+2^{n}\)，\(a_{1}=1\)，則\(a_{n}=2^{n}-1\)。（）答案：對10.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(S_{n}\)是其前\(n\)項和，若\(S_{10}=S_{20}\)，則\(S_{30}=0\)。（）答案：對四、簡答題1.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}=7\)，\(a_{5}+a_{7}=32\)，求\(\{a_{n}\}\)的通項公式。答案：設等差數(shù)列\(zhòng)(\{a_{n}\}\)的公差為\(d\)。已知\(a_{3}=7\)，則\(a_{1}+2d=7\)①；又\(a_{5}+a_{7}=32\)，即\(a_{1}+4d+a_{1}+6d=32\)，\(2a_{1}+10d=32\)，化簡得\(a_{1}+5d=16\)②。②-①得\(3d=9\)，\(d=3\)。把\(d=3\)代入①得\(a_{1}=1\)。所以\(a_{n}=a_{1}+(n-1)d=1+3(n-1)=3n-2\)。2.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=2\)，\(a_{4}=16\)，求數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)