总:381

高中数学选择性必修 第二册


单选题:


A.1

B.2

C.3

D.2017


参考答案:D


解析:


因为 \({a_n} + {a_{n + 1}} + {a_{n + 2}} = 2020\) ,


所以 \({a_{n + 1}} + {a_{n + 2}} + {a_{n + 3}} = 2020\) ,


两式相减可得, \({a_n} = {a_{n + 3}}\) ,


所以数列\(\left\{ {{a_n}} \right\}\)为以\(3\)为周期的周期数列,


又因为 \({a_{2020}} + {a_{2021}} + {a_{2022}} = 2020\) ,


即 \({a_{2020}} + 2 + 1 = 2020\) ,所以 \({a_{2020}} = 2017\) ,


因为 \(2020 = 673 \times 3 + 1\) ,所以 \({a_{2020}} = {a_1} = 2017\) .


故选:D



单选题:


A.\(2\)

B.\(3\)

C.\(6\)

D.\(8\)


参考答案:C


解析:


由题可知等积数列的各项以\(2\)​为一个周期循环出现,每相邻两项的和相等,前\(41\) 项的和为 \(103\) 


则 \(\left( {{a_1} + {a_2}} \right) + \left( {{a_3} + {a_4}} \right) + \cdots + \left( {{a_{39}} + {a_{40}}} \right) + {a_{41}} = 103\) 


即 \(20\left( {{a_1} + {a_2}} \right){\rm{ + }}{a_1} = 103\) ,解得 \({a_2} = 2\)  


所以公积是 \(2 \times 3{\rm{ = }}6\)  


故选C.



单选题:


A.1347

B.1348

C.1349

D.1346


参考答案:A


解析:


由数列\(1,1,2,3,5,8,13,21,34,…,\)​各项除以\(2\)​的余数,可得数列 \(\left\{ {{a_n}} \right\}\)​​ 为\(1,1,0,1,1,0,1,1,0,…,\)​所以数列\(\left\{ {{a_n}} \right\}\)是周期为3的周期数列,前三项和为 \(1 + 1 + 0 = 2\) ,又因为 \(2020 = 673 \times 3 + 1\) ,所以数列 \(\left\{ {{a_n}} \right\}\) 的前 2020 项的和为 \(673 \times 2 + 1 = 1347\) .


故选:A



单选题:


A.\(1840\)

B.\(1880\)

C.\(1960\)

D.\(1980\)


参考答案:A


解析:


由于\({a}_{n}={{n}^{2}}\left ( {\cos^{2} {\frac {n\pi } {3}-\sin^{2} {\frac {n\pi } {3}}}} \right )={{n}^{2}}\cos \frac {2n\pi } {3}\) ​​  


\(\therefore T = \frac{{2\pi }}{{\frac{{2\pi }}{3}}} = 3\) 


又 \({a_{3k - 2}} + {a_{3k - 1}} + {a_{3k}} = - \frac{1}{2}{(3k - 2)^2} - \frac{1}{2}{(3k - 1)^2} + {(3k)^2} = 9k - \frac{5}{2}\) 


\(\therefore {S_{60}} = 9({\rm{1 + 2 + }}...{\rm{ + 20)}} - \frac{5}{2} \times 20 = 1890 - 50 = 1840\) 


故选:A



单选题:


A.0

B.1

C.2

D.3


参考答案:C


解析:


由题公差 \(d > 0\) 的等差数列 \(\left\{ {{a_n}} \right\}\) ,设其通项公式 \({a_n} = {a_1} + \left( {n - 1} \right)d = dn + {a_1} - d\) ,是递增数列,(1)正确; \({a_n} + 3nd = 4dn + {a_1} - d\) 是递增数列,所以(4)正确;


若 \({a_n} = n - 10\) , \(n{a_n} = {n^2} - 10n\) 不是递增数列,所以(2)错误; \(\frac{{{a_n}}}{n} = \frac{{n - 10}}{n} = 1 - \frac{{10}}{n}\) 是递增数列,所以(3)错误.


故选:C



单选题:


A.107

B.108

C.\(108\frac{1}{8}\)

D.109


参考答案:B


解析:


由题意可知


\({a_n} = - 2{n^2} + 29n + 3 = - 2{\left( {n - \frac{{29}}{4}} \right)^2} + 108\frac{1}{8}\) ,


由于\(n \in {N^*}\),


故当\(n\)取距离 \(\frac{{29}}{4}\) 最近的正整数\(7\)时,\({a_n}\)取得最大值\(108\).


∴数列 \(\left\{ {{a_n}} \right\}\) 中的最大值为 \({a_7} = 108\) .


故选:B.



多选题:


A.5

B.6

C.7

D.8


参考答案:AB


解析:


当 \(n = 1\) 时, \({a_1} = {S_1} = \lambda {a_1} - 1\) , \(\therefore \lambda - 1 = 1\) ,解得: \(\lambda = 2\)


\(\therefore {S_n} = 2{a_n} - 1\) 


当 \(n \geqslant 2\) 且 \(n \in {N^ * }\) 时, \({S_{n - 1}} = 2{a_{n - 1}} - 1\) 


\(\therefore {a_n} = {S_n} - {S_{n - 1}} = 2{a_n} - 2{a_{n - 1}}\) ,即: \({a_n} = 2{a_{n - 1}}\) 


\(\therefore \)数列 \(\left\{ {{a_n}} \right\}\) 是以\(1\)为首项, \(2\) 为公比的等比数列, \(\therefore {a_n} = {2^{n - 1}}\) 


\(\because {a_n}{b_n} = - {n^2} + 9n - 20\) , \(\therefore {b_n} = \frac{{ - {n^2} + 9n - 20}}{{{2^{n - 1}}}}\) 


\(\therefore {b_{n + 1}} - {b_n} = \frac{{ - {{\left( {n + 1} \right)}^2} + 9\left( {n + 1} \right) - 20}}{{{2^n}}} - \frac{{ - {n^2} + 9n - 20}}{{{2^{n - 1}}}} = \frac{{{n^2} - 11n + 28}}{{{2^n}}} < 0\) 


\(\because {2^n} > 0\) , \(\therefore {n^2} - 11n + 28 = \left( {n - 4} \right)\left( {n - 7} \right) < 0\) ,解得: \(4 < n < 7\) 


又 \(n \in {N^ * }\) , \(\therefore n = 5\)或\(6\) 


故选:AB



题目:



参考答案:128


解析:


因为前 \(n\) 行一共有 \({n^2}\) 个数,且第 \(n\) 行的最后一个数为 \({n^2}\) ,


又因为 \({44^2} = 1936 < 2019,{45^2} = 2025 > 2019\) ,


所以 \(2019\) 在第 \(45\) 行,


且第 45 行最后数为 \({45^2} = 2025\) ,


又因为第 \(45\) 行有 \(2 \times 45 - 1 = 89\) 个数, \(2025 - 2019 = 6\) ,


所以 \(2019\) 在第 \(89 - 6 = 83\) 列,


所以\(i + j = 45 + 83 = 128\).


故答案为:\(128\).



题目:



参考答案:\(\frac{1}{2}\);29


解析:


由题意, \({a_1} = 2\) , \({a_2} = 3\) , \({a_n} = \frac{{{a_{n - 1}}}}{{{a_{n - 2}}}}\) ( \(n \in {{\rm{N}}^*}\) , \(n \geqslant 3\) ),


\(\therefore {a_3} = \frac{{{a_2}}}{{{a_1}}}{\rm{ = }}\frac{3}{2},{a_4} = \frac{{{a_3}}}{{{a_2}}}{\rm{ = }}\frac{1}{2},{a_5} = \frac{{{a_4}}}{{{a_3}}}{\rm{ = }}\frac{1}{3},{a_6} = \frac{{{a_5}}}{{{a_4}}}{\rm{ = }}\frac{2}{3},{a_7} = \frac{{{a_6}}}{{{a_5}}}{\rm{ = 2}},{a_8} = \frac{{{a_7}}}{{{a_6}}}{\rm{ = 3}},...\) 


故数列 \(\left\{ {{a_n}} \right\}\) 为周期 \(T = 6\) 的周期数列


\({a_{2020}} = {a_{336 \times 6 + 4}} = {a_4} = \frac{1}{2}\) 


由于 \({a_1} + {a_2} + {a_3} + {a_4} + {a_5} + {a_6} = 2 + 3 + \frac{3}{2} + \frac{1}{2} + \frac{1}{3} + \frac{2}{3} = 8\)  


故 \({S_{20}} = 3 \times 8{\rm{ + }}{a_1} + {a_2} = 29\) 


故答案为: \(\frac{1}{2}\),29



题目:



参考答案:\(10n - 2\);216


解析:


\({T_n}\) 为数列 \(\left\{ {{b_n}} \right\}\) 的前\(n\)项的和, \({T_n} = 5{n^2} + 3n\) ,


\({b_n} = {T_n} - {T_{n - 1}} = \left( {5{n^2} + 3n} \right) - \left[ {5{{\left( {n - 1} \right)}^2} + 3\left( {n - 1} \right)} \right] = 10n - 2\left( {n \geqslant 2} \right)\) 


验证 \(n = 1\) 时, \({b_1} = {T_1} = 8\) 也符合,故 \({b_n} = 10n - 2\) , \(\because {b_n} = {a_{{2^{n - 1}}}}\) , \(\therefore {a_{1024}} = {b_{11}} = 108\) ,\({a_{1025}} = 2{a_{1024}} = 216\) .


故答案为: \(10n - 2\) ; 216




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