已知函数  \(f(x) = \frac{1}{3}{x^3} + 4x\)  ，记等差数列  \(\left\{ {{a_n}} \right\}\)  的前\(n\)项和为 \({S_n}\) ，若\(f\left( {{a_1} + 2} \right) = 100\) ， \(f\left( {{a_{2022}} + 2} \right) = - 100\) ，则 \({S_{2022}} = \)（ ）

解：因为 \(f( - x) = - \frac{1}{3}{x^3} - 4x = - f(x),\therefore f(x)\) 是奇函数，

因为 \(f\left( {{a_1} + 2} \right) = 100\)，\(f\left( {{a_{2022}} + 2} \right) = - 100\)，所以 \(f\left( {{a_1} + 2} \right) = - f({a_{2022}} + 2)\)，

所以 \({a_1} + 2 + {a_{2022}} + 2 = 0\)，所以 \({a_1} + {a_{2022}} = - 4\)，

所以 \({S_{2022}} = \frac{{2022}}{2}({a_1} + {a_{2022}}) = - 4044\). 故选：A

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