數(shù)列關(guān)系試題及答案

一、單項(xiàng)選擇題1.數(shù)列\(zhòng)(1\)，\(3\)，\(6\)，\(10\)，\(15\)，…的一個(gè)通項(xiàng)公式是（）A.\(a_{n}=n^{2}-(n-1)\)B.\(a_{n}=n^{2}-1\)C.\(a_{n}=\frac{n(n+1)}{2}\)D.\(a_{n}=\frac{n(n-1)}{2}\)答案：C2.已知數(shù)列\(zhòng)(\{a_{n}\}\)的通項(xiàng)公式\(a_{n}=3n-50\)，則其前\(n\)項(xiàng)和\(S_{n}\)取得最小值時(shí)\(n\)的值為（）A.\(15\)B.\(16\)C.\(17\)D.\(18\)答案：B3.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}=5\)，\(a_{5}=9\)，則\(a_{7}\)等于（）A.\(12\)B.\(13\)C.\(14\)D.\(15\)答案：B4.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{2}=2\)，\(a_{5}=16\)，則公比\(q\)為（）A.\(-2\)B.\(2\)C.\(4\)D.\(8\)答案：B5.數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{1}=1\)，\(a_{n+1}=2a_{n}+1\)，則\(a_{5}\)的值為（）A.\(30\)B.\(31\)C.\(32\)D.\(33\)答案：B6.等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和為\(S_{n}\)，若\(S_{11}=22\)，則\(a_{6}\)等于（）A.\(1\)B.\(2\)C.\(3\)D.\(4\)答案：B7.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}+a_{2}=3\)，\(a_{3}+a_{4}=12\)，則\(a_{5}+a_{6}\)等于（）A.\(24\)B.\(48\)C.\(96\)D.\(192\)答案：B8.已知數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}=n^{2}+1\)，則\(a_{n}\)等于（）A.\(2n-1\)B.\(\begin{cases}2,&n=1\\2n-1,&n\geq2\end{cases}\)C.\(2n+1\)D.\(\begin{cases}2,&n=1\\2n+1,&n\geq2\end{cases}\)答案：B9.若數(shù)列\(zhòng)(\{a_{n}\}\)是等差數(shù)列，且\(a_{1}+a_{4}+a_{7}=45\)，\(a_{2}+a_{5}+a_{8}=39\)，則\(a_{3}+a_{6}+a_{9}\)的值是（）A.\(24\)B.\(27\)C.\(30\)D.\(33\)答案：D10.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}a_{4}a_{5}=8\)，則\(a_{2}a_{6}\)等于（）A.\(4\)B.\(6\)C.\(8\)D.\(16\)答案：A二、多項(xiàng)選擇題1.以下關(guān)于等差數(shù)列的說(shuō)法正確的是（）A.若\(\{a_{n}\}\)是等差數(shù)列，\(m,n,p,q\inN^+\)，且\(m+n=p+q\)，則\(a_{m}+a_{n}=a_{p}+a_{q}\)B.等差數(shù)列\(zhòng)(\{a_{n}\}\)的通項(xiàng)公式\(a_{n}=a_{1}+(n-1)d\)（\(d\)為公差）C.等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和公式\(S_{n}=\frac{n(a_{1}+a_{n})}{2}=na_{1}+\frac{n(n-1)}{2}d\)D.若\(a,b,c\)成等差數(shù)列，則\(2b=a+c\)答案：ABCD2.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，下列說(shuō)法正確的是（）A.若\(a_{1}>0\)，\(q>1\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是遞增數(shù)列B.若\(a_{1}<0\)，\(0<q<1\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是遞增數(shù)列C.若\(a_{n}=a_{1}q^{n-1}\)，則\(a_{m}=a_{n}q^{m-n}\)（\(m,n\inN^+\)）D.等比數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}=\frac{a_{1}(1-q^{n})}{1-q}\)（\(q\neq1\)）答案：ABCD3.已知數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}-a_{n}=2\)，\(a_{1}=1\)，則（）A.數(shù)列\(zhòng)(\{a_{n}\}\)是等差數(shù)列B.\(a_{n}=2n-1\)C.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}=n^{2}\)D.數(shù)列\(zhòng)(\{a_{n}\}\)是遞增數(shù)列答案：ABCD4.對(duì)于數(shù)列\(zhòng)(\{a_{n}\}\)，其前\(n\)項(xiàng)和\(S_{n}=3n^{2}-2n\)，則（）A.\(a_{1}=1\)B.\(a_{n}=6n-5\)C.數(shù)列\(zhòng)(\{a_{n}\}\)是等差數(shù)列D.數(shù)列\(zhòng)(\{a_{n}\}\)的公差為\(6\)答案：ABCD5.設(shè)等比數(shù)列\(zhòng)(\{a_{n}\}\)的公比為\(q\)，前\(n\)項(xiàng)和為\(S_{n}\)，若\(S_{n+1},S_{n},S_{n+2}\)成等差數(shù)列，則（）A.\(q=-2\)B.\(q=2\)C.\(a_{n+1}=-2a_{n}\)D.\(a_{n+1}=2a_{n}\)答案：AC6.數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(a_{n+1}=\frac{a_{n}}{1+a_{n}}\)，則（）A.\(a_{2}=\frac{1}{2}\)B.\(a_{3}=\frac{1}{3}\)C.\(a_{n}=\frac{1}{n}\)D.數(shù)列\(zhòng)(\{\frac{1}{a_{n}}\}\)是等差數(shù)列答案：ABCD7.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，公差\(d\neq0\)，\(a_{1}=1\)，且\(a_{1},a_{3},a_{9}\)成等比數(shù)列，則（）A.\(a_{n}=n\)B.\(a_{3}=3\)C.\(a_{9}=9\)D.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}=\frac{n(n+1)}{2}\)答案：ABCD8.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=2\)，\(a_{4}=16\)，則（）A.\(q=2\)B.\(a_{n}=2^{n}\)C.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}=2^{n+1}-2\)D.\(a_{2}=4\)答案：ABCD9.已知數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n}=2a_{n-1}+1\)（\(n\geq2\)），\(a_{1}=1\)，則（）A.\(a_{2}=3\)B.\(a_{3}=7\)C.\(a_{n}=2^{n}-1\)D.數(shù)列\(zhòng)(\{a_{n}\}\)是遞增數(shù)列答案：ABCD10.數(shù)列\(zhòng)(\{a_{n}\}\)的通項(xiàng)公式\(a_{n}=n^{2}-9n\)，則（）A.數(shù)列\(zhòng)(\{a_{n}\}\)的最小項(xiàng)是\(a_{4}\)或\(a_{5}\)B.\(a_{4}=-20\)C.\(a_{5}=-20\)D.數(shù)列\(zhòng)(\{a_{n}\}\)在\(n<4.5\)時(shí)單調(diào)遞減，在\(n>4.5\)時(shí)單調(diào)遞增答案：ABCD三、判斷題1.常數(shù)列既是等差數(shù)列又是等比數(shù)列。（×）2.若數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}=n^{2}+n+1\)，則\(a_{n}=2n\)。（×）3.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{m}=a_{n}\)（\(m\neqn\)），則公差\(d=0\)。（√）4.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}+a_{2}\neq0\)，若\(a_{1},a_{3},a_{2}\)成等差數(shù)列，則公比\(q=-\frac{1}{2}\)。（√）5.若數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}-a_{n}=n\)，則\(a_{n}\)的通項(xiàng)公式可通過(guò)累加法求得。（√）6.等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}\)，\(S_{m},S_{2m}-S_{m},S_{3m}-S_{2m}\)（\(m\inN^+\)）也成等差數(shù)列。（√）7.等比數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}\)，當(dāng)\(q=-1\)且\(n\)為偶數(shù)時(shí)，\(S_{n}=0\)。（√）8.若數(shù)列\(zhòng)(\{a_{n}\}\)的通項(xiàng)公式\(a_{n}=(-1)^{n}n\)，則\(S_{100}=50\)。（√）9.數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{n+1}=a_{n}+2\)（\(n\geq1\)），則\(\{a_{n}\}\)是公差為\(2\)的等差數(shù)列。（√）10.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{n}>0\)，\(a_{2}a_{4}+2a_{3}a_{5}+a_{4}a_{6}=25\)，則\(a_{3}+a_{5}=5\)。（√）四、簡(jiǎn)答題1.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}=7\)，\(a_{5}+a_{7}=26\)，求\(\{a_{n}\}\)的通項(xiàng)公式。答：設(shè)等差數(shù)列\(zhòng)(\{a_{n}\}\)的公差為\(d\)，首項(xiàng)為\(a_{1}\)。已知\(a_{3}=7\)，則\(a_{1}+2d=7\)；\(a_{5}+a_{7}=26\)，即\((a_{1}+4d)+(a_{1}+6d)=26\)，化簡(jiǎn)得\(2a_{1}+10d=26\)。聯(lián)立方程組\(\begin{cases}a_{1}+2d=7\\2a_{1}+10d=26\end{cases}\)，解得\(a_{1}=3\)，\(d=2\)。所以\(a_{n}=a_{1}+(n-1)d=3+2(n-1)=2n+1\)。2.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}=12\)，\(a_{4}=18\)，求\(\{a_{n}\}\)的通項(xiàng)公式。答：設(shè)等比數(shù)列\(zhòng)(\{a_{n}\}\)的公比為\(q\)。由等比數(shù)列性質(zhì)可知\(q=\frac{a_{4}}{a_{3}}=\frac{18}{12}=\frac{3}{2}\)。又\(a_{3}=a_{1}q^{2}=12\)，把\(q=\frac{3}{2}\)代入可得\(a_{1}\times(\frac{3}{2})^{2}=12\)，解得\(a_{1}=\frac{16}{3}\)。所以通項(xiàng)公式\(a_{n}=a_{1}q^{n-1}=\frac{16}{3}\times(\frac{3}{2})^{n-1}=2^{n+1}\times3^{2-n}\)。3.已知數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}=n^{2}-2n\)，求\(a_{n}\)。答：當(dāng)\(n=1\)時(shí)，\(a_{1}=S_{1}=1^{2}-2\times1=-1\)。當(dāng)\(n\geq2\)時(shí)，\(a_{n}=S_{n}-S_{n-1}=n^{2}-2n-[(n-1)^{2}-2(n-1)]\)，化簡(jiǎn)得\(a_{n}=n^{2}-2n-(n^{2}-2n+1-2n+2)=2n-3\)。當(dāng)\(n=1\)時(shí)，\(2\times1-3=-1=a_{1}\)。所以\(a_{n}=2n-3\)。4.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(d=2\)，求其前\(n\)項(xiàng)和\(S_{n}\)的最小值。答：由等差數(shù)列前\(n\)項(xiàng)和公式\(S_{n}=na_{1}+\frac