已知数列 \(\left\{ {{a_n}} \right\}\) 各项均不为零，且 \({a_1} = 3,\frac{{{a_{n + 1}}}}{{{a_{n + 1}} - {a_n}}} - \frac{{{a_{n - 1}}}}{{{a_n} - {a_{n - 1}}}} = 2\left( {n \geqslant 2,n \in {N^*}} \right)\) ，若 \({a_{20}} = 41\) ，则 \({a_9} = \) （       ）

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.

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