数列 \(\left\{ {{a_n}} \right\}\) 中，前 \(n\) 项和为 \({S_n}\) ．若 \({a_1} = 2\) ， \({a_2} = 3\) ， \({a_n} = \frac{{{a_{n - 1}}}}{{{a_{n - 2}}}}\) （ \(n \in {{\rm{N}}^*}\) ， \(n \geqslant 3\) ），则 \({a_{2020}} = \) ___； ___．

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

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