已知 $a,b$ 为实数，满足 ${\left( {a + b} \right)^{59}} = - 1$，${\left( {a - b} \right)^{60}} = 1$，则 $\displaystyle \sum\limits_{n = 1}^{60} {\left({{a^n} - {b^n}} \right)} =$ \((\qquad)\)
【难度】
【出处】
2007年复旦大学优秀高中生文化水平选拔测试
【标注】
【答案】
C
【解析】
由$$\begin{cases}
a + b = - 1 ,\\
a - b = 1 .\\
\end{cases}$$解得$$\begin{cases}a = 0 ,\\
b = - 1 ;\\
\end{cases}$$由$$\begin{cases}a + b = - 1 ,\\
a - b = - 1 .\\
\end{cases}$$解得$$\begin{cases}a = - 1 ,\\
b = 0 .\\
\end{cases}$$所以$$ \sum\limits_{n = 1}^{60} {\left( {{a^n} - {b^n}} \right)} = \sum\limits_{n = 1}^{60} {{a^n}} - \sum\limits_{n = 1}^{60} {{b^n}} = 0.$$
a + b = - 1 ,\\
a - b = 1 .\\
\end{cases}$$解得$$\begin{cases}a = 0 ,\\
b = - 1 ;\\
\end{cases}$$由$$\begin{cases}a + b = - 1 ,\\
a - b = - 1 .\\
\end{cases}$$解得$$\begin{cases}a = - 1 ,\\
b = 0 .\\
\end{cases}$$所以$$ \sum\limits_{n = 1}^{60} {\left( {{a^n} - {b^n}} \right)} = \sum\limits_{n = 1}^{60} {{a^n}} - \sum\limits_{n = 1}^{60} {{b^n}} = 0.$$
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