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