高中數(shù)列遞增題目及答案一、單項選擇題1.已知數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}-a_{n}=2\)，\(a_{1}=1\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是（）A.遞增數(shù)列B.遞減數(shù)列C.常數(shù)列D.擺動數(shù)列答案：A2.在等差數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}<0\)，\(d>0\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是（）A.遞增數(shù)列B.遞減數(shù)列C.常數(shù)列D.擺動數(shù)列答案：A3.已知等比數(shù)列\(zhòng)(\{a_{n}\}\)的公比\(q>1\)，\(a_{1}>0\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是（）A.遞增數(shù)列B.遞減數(shù)列C.常數(shù)列D.擺動數(shù)列答案：A4.已知數(shù)列\(zhòng)(\{a_{n}\}\)的通項公式為\(a_{n}=n^{2}-5n+4\)，則數(shù)列\(zhòng)(\{a_{n}\}\)的單調(diào)遞增區(qū)間是（）A.\((\frac{5}{2},+\infty)\)B.\((2,+\infty)\)C.\((-\infty,\frac{5}{2})\)D.\((-\infty,2)\)答案：A5.已知數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}=\frac{a_{n}}{1+a_{n}}\)，\(a_{1}=1\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是（）A.遞增數(shù)列B.遞減數(shù)列C.常數(shù)列D.擺動數(shù)列答案：B6.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{3}<0\)，\(a_{4}>0\)，則數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}\)的最小值是（）A.\(S_{3}\)B.\(S_{4}\)C.\(S_{3}\)或\(S_{4}\)D.無法確定答案：A7.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}>0\)，\(0<q<1\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是（）A.遞增數(shù)列B.遞減數(shù)列C.常數(shù)列D.擺動數(shù)列答案：B8.已知數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=n^{2}-2n+1\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是（）A.遞增數(shù)列B.遞減數(shù)列C.常數(shù)列D.擺動數(shù)列答案：A9.已知數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}=a_{n}+\frac{1}{n(n+1)}\)，\(a_{1}=1\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是（）A.遞增數(shù)列B.遞減數(shù)列C.常數(shù)列D.擺動數(shù)列答案：A10.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}>0\)，\(d<0\)，且\(\verta_{3}\vert=\verta_{9}\vert\)，則數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}\)取得最大值時\(n\)的值為（）A.5B.6C.5或6D.6或7答案：C二、多項選擇題1.下列數(shù)列中是遞增數(shù)列的有（）A.\(a_{n}=2n-3\)B.\(a_{n}=3^{n}\)C.\(a_{n}=(\frac{1}{2})^{n}\)D.\(a_{n}=n^{2}-2n\)答案：AB2.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}<0\)，\(d>0\)，則下列結(jié)論正確的有（）A.數(shù)列\(zhòng)(\{a_{n}\}\)單調(diào)遞增B.數(shù)列\(zhòng)(\{a_{n}\}\)單調(diào)遞減C.當\(n\geq2\)時，\(a_{n}>0\)D.\(S_{n}\)有最小值答案：AD3.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}>0\)，\(q>1\)，則下列結(jié)論正確的有（）A.數(shù)列\(zhòng)(\{a_{n}\}\)單調(diào)遞增B.數(shù)列\(zhòng)(\{a_{n}\}\)單調(diào)遞減C.當\(n\geq2\)時，\(a_{n}>a_{1}\)D.\(S_{n}\)有最小值答案：AC4.已知數(shù)列\(zhòng)(\{a_{n}\}\)的通項公式為\(a_{n}=n^{2}-kn+2\)，若數(shù)列\(zhòng)(\{a_{n}\}\)單調(diào)遞增，則\(k\)的取值可以是（）A.1B.2C.3D.4答案：ABC5.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(S_{n}\)為其前\(n\)項和，若\(S_{10}<0\)，\(S_{11}>0\)，則下列結(jié)論正確的有（）A.\(a_{1}<0\)B.\(d>0\)C.\(a_{6}>0\)D.\(S_{n}\)的最小值為\(S_{6}\)答案：ABCD三、判斷題1.若數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}>a_{n}\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是遞增數(shù)列。（）答案：√2.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(d>0\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是遞增數(shù)列。（）答案：√3.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(q>1\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是遞增數(shù)列。（）答案：×4.已知數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=n^{2}-1\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是遞增數(shù)列。（）答案：×5.若數(shù)列\(zhòng)(\{a_{n}\}\)的通項公式為\(a_{n}=(-1)^{n}\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是遞增數(shù)列。（）答案：×6.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}>0\)，\(S_{9}=S_{16}\)，則\(S_{n}\)取得最大值時\(n=12\)或\(13\)。（）答案：√7.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}<0\)，\(0<q<1\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是遞增數(shù)列。（）答案：√8.已知數(shù)列\(zhòng)(\{a_{n}\}\)的通項公式為\(a_{n}=2n-1\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是遞增數(shù)列。（）答案：√9.若數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}-a_{n}=1\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是遞增數(shù)列。（）答案：√10.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}<0\)，\(S_{10}>0\)，則\(S_{9}<0\)。（）答案：√四、簡答題1.簡述等差數(shù)列\(zhòng)(\{a_{n}\}\)遞增的條件。答案：等差數(shù)列\(zhòng)(\{a_{n}\}\)遞增的條件是公差\(d>0\)，即后一項比前一項大。2.簡述等比數(shù)列\(zhòng)(\{a_{n}\}\)遞增的條件。答案：等比數(shù)列\(zhòng)(\{a_{n}\}\)遞增的條件有兩種情況。當\(a_{1}>0\)且\(q>1\)時，數(shù)列遞增；當\(a_{1}<0\)且\(0<q<1\)時，數(shù)列也遞增。3.已知數(shù)列\(zhòng)(\{a_{n}\}\)的通項公式為\(a_{n}=2n-1\)，證明數(shù)列\(zhòng)(\{a_{n}\}\)是遞增數(shù)列。答案：對于\(a_{n}=2n-1\)，\(a_{n+1}-a_{n}=[2(n+1)-1]-(2n-1)=2>0\)，所以\(a_{n+1}>a_{n}\)，數(shù)列\(zhòng)(\{a_{n}\}\)是遞增數(shù)列。4.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=2\)，\(d=3\)，求數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}\)的最小值。答案：由\(a_{n}=a_{1}+(n-1)d=2+3(n-1)=3n-1\)。令\(a_{n}\leq0\)，即\(3n-1\leq0\)，解得\(n\leq\frac{1}{3}\)，因為\(n\)為正整數(shù)，所以當\(n=1\)時，\(a_{n}\)最小為\(2\)。\(S_{n}=na_{1}+\frac{n(n-1)}{2}d=2n+\frac{3n(n-1)}{2}=\frac{3}{2}n^{2}+\frac{1}{2}n\)，當\(n=1\)時，\(S_{n}\)最小為\(\frac{3}{2}+\frac{1}{2}=2\)。五、討論題1.討論等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(S_{n}\)與\(S_{2n}-S_{n}\)的大小關(guān)系。答案：設(shè)等差數(shù)列\(zhòng)(\{a_{n}\}\)的公差為\(d\)，\(S_{n}=na_{1}+\frac{n(n-1)}{2}d\)，\(S_{2n}=2na_{1}+\frac{2n(2n-1)}{2}d\)，則\(S_{2n}-S_{n}=2na_{1}+\frac{2n(2n-1)}{2}d-(na_{1}+\frac{n(n-1)}{2}d)=na_{1}+\frac{n(3n-1)}{2}d\)。\(S_{2n}-S_{n}-S_{n}=na_{1}+\frac{n(3n-1)}{2}d-(na_{1}+\frac{n(n-1)}{2}d)=\frac{n(2n)}{2}d=n^{2}d\)。當\(d>0\)時，\(n^{2}d>0\)，即\(S_{2n}-S_{n}>S_{n}\)；當\(d=0\)時，\(n^{2}d=0\)，即\(S_{2n}-S_{n}=S_{n}\)；當\(d<0\)時，\(n^{2}d<0\)，即\(S_{2n}-S_{n}<S_{n}\)。2.討論等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(S_{n}\)與\(S_{2n}-S_{n}\)的大小關(guān)系。答案：設(shè)等比數(shù)列\(zhòng)(\{a_{n}\}\)的公比為\(q\)，首項為\(a_{1}\)。當\(q=1\)時，\(S_{n}=na_{1}\)，\(S_{2n}=2na_{1}\)，則\(S_{2n}-S_{n}=2na_{1}-na_{1}=na_{1}\)，所以當\(a_{1}>0\)時，\(S_{2n}-S_{n}>S_{n}\)；當\(a_{1}<0\)時，\(S_{2n}-S_{n}<S_{n}\)。當\(q\neq1\)時，\(S_{n}=\frac{a_{1}(1-q^{n})}{1-q}\)，\(S_{2n}=\frac{a_{1}(1-q^{2n})}{1-q}\)，則\(S_{2n}-S_{n}=\frac{a_{1}(1-q^{2n})}{1-q}-\frac{a_{1}(1-q^{n})}{1-q}=\frac{a_{1}(q^{n}-q^{2n})}{1-q}=\frac{a_{1}q^{n}(1-q^{n})}{1-q}\)。當\(q>1\)或\(q<0\)時，若\(a_{1}>0\)且\(1-q^{n}>0\)，則\(S_{2n}-S_{n}>S_{n}\)；若\(a_{1}<0\)且\(1-q^{n}<0\)，則\(S_{2n}-S_{n}<S_{n}\)。當\(0<q<1\)時，若\(a_{1}>0\)且\(1-q^{n}<0\)，則\(S_{2n}-S_{n}>S_{n}\)；若\(a_{1}<0\)且\(1-q^{n}>0\)，則\(S_{2n}-S_{n}<S_{n}\)。3.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和為\(S_{n}\)，且\(S_{10}=100\)，\(S_{100}=10\)，求\(S_{110}\)。答案：設(shè)等差數(shù)列\(zhòng)(\{a_{n}\}\)的公差為\(d