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

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.

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

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

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

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.

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

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：\(128\).

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

\({T_n}\) 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

已知单调数列 \(\left\{ {{a_n}} \right\}\) 的前 \(n\) 项和为 \({S_n}\) ，若 \({S_n} + {S_{n + 1}} = {n^2} + n\) ，则首项 \({a_1}\) 的取值范围是___．

当 \(n = 1\) 时， \({S_1} + {S_2} = 2\) ， ∴ \({a_2} = 2 - 2{a_1}\)，

当 \({\rm{ }}n \geqslant 2\) 时 \({S_n} + {S_{n + 1}} = {n^2} + n\) ，

\({S_{n - 1}} + {S_n} = {(n - 1)^2} + (n - 1)\) ，

两式相减得 \({a_n} + {a_{n + 1}} = 2n\) ①． \({a_2} + {a_3} = 4\) ， \({a_3} = 2 + 2{a_1}\) ，

当 \(n \geqslant 3{\rm{ }}\) 时， \({a_{n - 1}} + {a_n} = 2(n - 1)\) ②，

 \(① - ②\) 得 \({a_{n + 1}} - {a_{n - 1}} = 2\) ，

∴数列 \(\left\{ {{a_n}} \right\}\) 从第2项起，偶数项成公差为\(2\)的等差数列，从第3项起，奇数项成公差为\(2\)的等差数列

∴数列 \(\left\{ {{a_n}} \right\}\) 单调递增，则满足 \({a_1} < {a_2} < {a_3} < {a_2} + 2\) ， ∴ \({a_1} < 2 - 2{a_1} < 2 + 2{a_1} < 4 - 2{a_1}\)，解得 \(0 < {a_1} < \frac{1}{2}\) ．

因为第 63 行的第一个数是：

\(1 + 63 \times (63 - 1) \div 2 = 1954\)  

而 \(2013 - 1954 = 59\)  

所以 \(59{\rm{ + }}1{\rm{ = }}60\)  

数字 \(2013\) 是第 \(63\) 行左起第 \(60\) 个数

即 \(m = 63,n = 60\)  

故 \(\frac{n}{m} = \frac{{60}}{{63}} = \frac{{20}}{{21}}\)  

故选：B

已知数列 \(1,\frac{1}{2},\frac{1}{{{2^2}}},\frac{3}{{{2^2}}},\frac{1}{{{2^3}}},\frac{3}{{{2^3}}},\frac{5}{{{2^3}}}\frac{7}{{{2^3}}},...,\frac{1}{{{2^n}}},\frac{3}{{{2^n}}},\frac{5}{{{2^n}}},...\) ，则该数列第\(2019\)项是（       ）

由数列\(\left( 1 \right),\left( {\frac{1}{2}} \right),\left( {\frac{1}{{{2^2}}},\frac{3}{{{2^2}}}} \right)\left( {\frac{1}{{{2^3}}},\frac{3}{{{2^3}}},\frac{5}{{{2^3}}},\frac{7}{{{2^3}}}} \right),...,\left( {\frac{1}{{{2^n}}},\frac{3}{{{2^n}}},\frac{5}{{{2^n}}},...} \right)\)，可发现其项数为

\(1,1,2,4,8,...,{2^{k - 2}},...\)，则前\(12\)个括号里共有\(1024\)项，前\(13\)个括号里共有\(2048\)项，

故原数列第\(2019\)项是第\(13\)个括号里的第\(995\)项，第\(13\)个括号里的数列通项为\(\frac{{2m - 1}}{{{2^{11}}}}\)，

所以第\(12\)个括号里的第\(995\)项是\(\frac{{1989}}{{{2^{11}}}}\).

故选：C.

（1） \(\because 0 \leqslant {a_1} \leqslant 1\) ，定义 \({a_{n + 1}} = \left\{ {\begin{array}{*{20}{l}} {2{a_n},0 \leqslant {a_n} < \frac{1}{2}} \\\ {2{a_n} - 1,{a_n} \geqslant \frac{1}{2}} \end{array}} \right.\).

若 \(0 \leqslant {a_2} < \frac{1}{2}\) ，则 \({a_3} = 2{a_2} = {a_2}\) ，解得 \({a_2} = 0\)；

若 \({a_2} \geqslant \frac{1}{2}\) ，则 \({a_3} = 2{a_2} - 1 = {a_2}\) ，解得 \({a_2} = 1\) .

综上所述， \({a_2} = 0\) 或 \(1\) ；

（2）①当 \(0 \leqslant {a_1} < \frac{1}{2}\) 时， \({a_2} = 2{a_1}\) .

（i）若 \(0 \leqslant {a_2} < \frac{1}{2}\) ，即 \(0 \leqslant 2{a_1} < \frac{1}{2} \Rightarrow 0 \leqslant {a_1} < \frac{1}{4}\) ， \({a_3} = 2{a_2} = 4{a_1}\) ，

\(\because {a_1} < {a_3} = 4{a_1}\) ， \(\therefore {a_1} > 0\) ，此时， \(0 < {a_1} < \frac{1}{4}\) ；

（ii）若 \({a_2} \geqslant \frac{1}{2}\) ，即 \(2{a_1} \geqslant \frac{1}{2}\) ，得 \(\frac{1}{4} \leqslant {a_1} < \frac{1}{2}\) ， \({a_3} = 2{a_2} - 1 = 4{a_1} - 1\) ，

\(\because {a_1} < {a_3} = 4{a_1} - 1\) ， \(\therefore {a_1} > \frac{1}{3}\) ，此时， \(\frac{1}{3} < {a_1} < \frac{1}{2}\) ；

②当 \(\frac{1}{2} \leqslant {a_1} \leqslant 1\) 时， \({a_2} = 2{a_1} - 1\) .

（i）若 \(0 \leqslant {a_2} < \frac{1}{2}\) ，即 \(0 \leqslant 2{a_1} - 1 < \frac{1}{2} \Rightarrow \frac{1}{2} \leqslant {a_1} < \frac{3}{4}\) ， \({a_3} = 2{a_2} = 4{a_1} - 2\) ，

 \(\because {a_1} < {a_3} = 4{a_2} - 2\) ， \(\therefore {a_1} > \frac{2}{3}\) ，此时， \(\frac{2}{3} < {a_1} < \frac{3}{4}\) ；

（ii）若 \({a_2} \geqslant \frac{1}{2}\) ，即 \(2{a_1} - 1 \geqslant \frac{1}{2} \Rightarrow \frac{3}{4} \leqslant {a_1} \leqslant 1\) ， \({a_3} = 2{a_2} - 1 = 4{a_1} - 3\) ，

\(\because {a_1} < {a_3} = 4{a_1} - 3\) ， \(\therefore {a_1} > 1\) ，此时， \({a_1}\) 不存在.

综上所述， \({a_1}\) 的取值范围是 \(\left( {0,\frac{1}{4}} \right) \cup \left( {\frac{1}{3},\frac{1}{2}} \right) \cup \left( {\frac{2}{3},\frac{3}{4}} \right)\) .

故答案为： \(0\) 或 \(1\) ； \(\left( {0,\frac{1}{4}} \right) \cup \left( {\frac{1}{3},\frac{1}{2}} \right) \cup \left( {\frac{2}{3},\frac{3}{4}} \right)\) .

由 \(\frac{{{a_{n + 1}}}}{{{a_{n + 1}} - {a_n}}} - \frac{{{a_{n - 1}}}}{{{a_n} - {a_{n - 1}}}} = 2\left( {n \geqslant 2,n \in {N^*}} \right)\) 得 ， \(\frac{{{a_{n + 1}} - {a_n} + {a_n}}}{{{a_{n + 1}} - {a_n}}} - \frac{{{a_{n - 1}}}}{{{a_n} - {a_{n - 1}}}} = 2\)

则 \(\frac{{{a_n}}}{{{a_{n + 1}} - {a_n}}} - \frac{{{a_{n - 1}}}}{{{a_n} - {a_{n - 1}}}} = 1,n \geqslant 2\) .

令 \(\frac{{{a_1}}}{{{a_2} - {a_1}}} = t + 1\) ，则数列 \(\left\{ {\frac{{{a_n}}}{{{a_{n + 1}} - {a_n}}}} \right\}\) 是公差为1，首项为\(t+1\)的等差数列，所以\(\frac{{{a_n}}}{{{a_{n + 1}} - {a_n}}} = n + t\) ，所以 \(\frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{{n + t + 1}}{{n + t}}\) .

所以 \({a_n} = {a_1} \times \frac{{{a_2}}}{{{a_1}}} \times \frac{{{a_3}}}{{{a_2}}} \times \cdots \times \frac{{{a_n}}}{{{a_{n - 1}}}} = 3 \times \frac{{2 + t}}{{1 + t}} \times \frac{{3 + t}}{{2 + t}} \times \cdots \times \frac{{n + t}}{{n + t - 1}} = \frac{{3n + 3t}}{{1 + t}}(n \geqslant 2)\) 

当 \(n=1\) 时， \({a_1} = 3\) ，也符合上式，所以 \({a_n} = \frac{{3n + 3t}}{{1 + t}}\left( {n \in {N^*}} \right)\) ；

所以 \({a_{20}} = \frac{{60 + 3t}}{{1 + t}} = 41\) ,解得 \(t = \frac{1}{2}\) ,

所以 \({a_n} = \frac{{3n + 3 \times \frac{1}{2}}}{{1 + \frac{1}{2}}}{\rm{ = 2}}n + 1\left( {n \in {N^*}} \right)\) ，

所以 \({a_9} = 19\) ，

故选A.

因 \({a_n} = {2^n}\) ， \({b_n} = {2^{{a_n}}} = {2^{{2^n}}}\) ， \(\left( {1 + \frac{1}{{{b_1}}}} \right)\left( {1 + \frac{1}{{{b_2}}}} \right)\left( {1 + \frac{1}{{{b_3}}}} \right) \cdots \left( {1 + \frac{1}{{{b_n}}}} \right)\) 

\( = \left( {1 + \frac{1}{{{2^2}}}} \right)\left( {1 + \frac{1}{{{2^{{2^2}}}}}} \right)\left( {1 + \frac{1}{{{2^{{2^3}}}}}} \right) \cdots \left( {1 + \frac{1}{{{2^{{2^n}}}}}} \right)\) \( = \frac{{\left( {1 - \frac{1}{{{2^2}}}} \right)\left( {1 + \frac{1}{{{2^2}}}} \right)\left( {1 + \frac{1}{{{2^{{2^2}}}}}} \right)\left( {1 + \frac{1}{{{2^{{2^3}}}}}} \right) \cdots \left( {1 + \frac{1}{{{2^{{2^n}}}}}} \right)}}{{1 - \frac{1}{{{2^2}}}}}\) 

\( = \frac{{\left( {1 - {{\left( {\frac{1}{{{2^{{2^n}}}}}} \right)}^2}} \right)}}{{1 - \frac{1}{{{2^2}}}}} < \frac{4}{3} < 2\) .

故选：C.

数列 \({a_1},{a_2} - {a_1},{a_3} - {a_2}, \ldots ,{a_n} - {a_n}_{ - 1}\) ，…是首项为\(1\)，公比为\(2\)的等比数列，那么\({a_n} = \) ___.

\({a_n} - {a_n}_{ - 1} = {a_1}{q^n}^{ - 1} = {2^n}^{ - 1}\) ，

即 \(\left\{ {\begin{array}{*{20}{l}}
{{a_2} - {a_1} = 2,} \\\
{{a_3} - {a_2} = {2^2},} \\\
{...} \\\
{{a_n} - {a_{n - 1}} = {2^{n - 1}}.}
\end{array}} \right.\) 

各式相加得 \({a_n} - {a_1} = 2 + {2^2} + \ldots + {2^n}^{ - 1} = {2^n} - 2\) ，

故 \({a_n} = {a_1} + {2^n} - 2 = {2^n} - 1\) .

故答案为： \({2^n} - 1\)

已知数列 \(\left\{ {{a_n}} \right\}\) 的前\(n\)项和为 \({S_n}\) ，且 \({S_n} = 2{a_n} - 1\) ，则数列 \(\left\{ {{a_n}} \right\}\) 的通项公式\({a_n} = \)___．

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

当 \(n \geqslant 2\) 时， \({a_n} = {S_n} - {S_{n - 1}} = 2{a_n} - 1 - \left( {2{a_{n - 1}} - 1} \right)\) ， \(\therefore {a_n} = 2{a_{n - 1}}\) ，

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

故答案为：\({2^{n - 1}}\).

已知数列 \(\{ {a_n}\} \) 满足： \({a_1} = 1\) ，且 \((n + 1){a_{n + 1}} - n{a_n} - 3 = 0\) ，若对任意的 \(a \in [ - 2,2]\) ，不等式 \({a_n} \leqslant 2{t^2} + at - 1\) 恒成立，则实数 \(t\) 的范围为___

由 \((n + 1){a_{n + 1}} - n{a_n} - 3 = 0\) 得， \((n + 1){a_{n + 1}} - n{a_n} = 3\) ，所以数列 \(\{ n{a_n}\} \) 是以 1 为首项，3为公差的等差数列．所以 \(n{a_n} = 1 + 3\left( {n - 1} \right) = 3n - 2\) ，即 \({a_n} = 3 - \frac{2}{n} < 3\) ，

因为 \(g(a) = at + 2{t^2} - 1\) 在 \(a \in [ - 2,2]\) 上单调，所以 \({g_{\min }} = \min \left\{ {g\left( { - 2} \right),g\left( 2 \right)} \right\}\) ，

因此可得 \(\left\{ {\begin{array}{*{20}{c}}
{g\left( { - 2} \right) \geqslant 3} \\\
{g\left( 2 \right) \geqslant 3}
\end{array}} \right.\) 即 \(\left\{ {\begin{array}{*{20}{c}}
{ - 2t + 2{t^2} - 1 \geqslant 3} \\\
{2t + 2{t^2} - 1 \geqslant 3}
\end{array}} \right.\) ，解得 \(t \geqslant 2\) 或 \(t \leqslant - 2\) ．

故答案为 \(t \geqslant 2\) 或 \(t \leqslant - 2\) ．

著名的斐波那契数列： 1，1，2，3，5，…， 的特点是从三个数起，每一个数等于它前面两个数的和，则 \(\frac{{a_1^2 + a_2^2 + a_3^2 + \cdots + a_{2048}^2}}{{{a_{2048}}}}\) 是数列中的第___项．

由题意可知 \({a_{n + 2}} = {a_{n + 1}} + {a_n}\) ，

所以 \({a_{n + 1}} \cdot {a_{n + 2}} = {a_{n + 1}} \cdot \left( {{a_{n + 1}} + {a_n}} \right)\) ，即 \({a_{n + 1}} \cdot {a_{n + 2}} = a_{n + 1}^2 + {a_{n + 1}} \cdot {a_n}\) 

所以 \({a_{2048}} \cdot {a_{2049}} = a_{2048}^2 + {a_{2048}} \cdot {a_{2047}}\) ， 

\({a_{2047}} \cdot {a_{2048}} = a_{2047}^2 + {a_{2047}} \cdot {a_{2046}}\) ， 

……

\({a_2} \cdot {a_3} = a_2^2 + {a_2}\cdot{a_1}\) ， 

所以 \({a_{2048}} \cdot {a_{2049}} = a_{2048}^2 + a_{2047}^2 + \ldots + a_2^2 + {a_2}\cdot{a_1}\) ，

又 \({a_2} = {a_1}\)  

所以 \({a_{2048}} \cdot {a_{2049}} = a_{2048}^2 + a_{2047}^2 + \ldots + a_2^2 + a_1^2\)  

∴  \(\frac{{a_1^2 + a_2^2 + a_3^2 + \cdots + a_{2048}^2}}{{{a_{2048}}}} = {a_{2049}}\) ． 

所以 \(\frac{{a_1^2 + a_2^2 + a_3^2 + \cdots + a_{2048}^2}}{{{a_{2048}}}}\) 是数列中的第\(2049\)项.

故答案为：\(2049\) ．