参考答案:\({a_n} = \frac{1}{3}{\left( {\frac{5}{4}} \right)^{n - 1}} + \frac{2}{3}\)
由题意可知,\({a_1} = 1\),第 \(n + 1\) 年后, \({a_{n + 1}} = \left( {1 + 25\% } \right){a_n} - \frac{1}{6} = \frac{5}{4}{a_n} - \frac{1}{6}\) ,
则 \({a_{n + 1}} - \frac{2}{3} = \frac{5}{4}\left( {{a_n} - \frac{2}{3}} \right)\) ,所以,数列 \(\left\{ {{a_n} - \frac{2}{3}} \right\}\) 是以 \(\frac{1}{3}\) 为首项,以 \(\frac{5}{4}\)为公比的等比数列,
则 \({a_n} - \frac{2}{3} = \frac{1}{3} \cdot {\left( {\frac{5}{4}} \right)^{n - 1}}\) ,因此, \({a_n} = \frac{1}{3}{\left( {\frac{5}{4}} \right)^{n - 1}} + \frac{2}{3}\) .
故答案为: \({a_n} = \frac{1}{3}{\left( {\frac{5}{4}} \right)^{n - 1}} + \frac{2}{3}\) .