已知数列 \(\left\{ {{a_n}} \right\}\) 的通项 \({a_n} = {n^2}\left( {{{\cos }^2}\frac{{n\pi }}{3} - {{\sin }^2}\frac{{n\pi }}{3}} \right)\) ，其前 \(n\) 项和为 \({S_n}\) ，则 \({{S}_{60}}=\)（       ）

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

进入考试题库