设 ${S_n}$ 为数列 $ \left\{ {a_n} \right\}$ 的前 $ n $ 项和，${S_n} = {\left( - 1\right)^n}{a_n} - \dfrac{1}{2^n},n \in {{\mathbb{N}}^ * } $，则 ${a_3} = $_______；${S_1} + {S_2} + \cdot \cdot \cdot + {S_{100}} = $_______．
答案 $-\dfrac 1 {16}$；$\dfrac 1 3 \left(\dfrac 1 {2^{100}} -1\right) $．
解析 根据题意，有\[S_n=\begin{cases} -a_n-\dfrac1{2^n},&n~\text{是奇数},\\ a_n-\dfrac1{2^n},&n~\text{是偶数},\end{cases}\iff \begin{cases} S_n=\dfrac 12S_{n-1}-\dfrac{1}{2^{n+1}},&n~\text{是奇数},\\ S_{n-1}=-\dfrac{1}{2^n},&n~\text{是偶数},\end{cases}\]于是\[S_n=\begin{cases} 0,&n~\text{是偶数},\\ -\dfrac{1}{2^{n+1}},&n~\text{是奇数},\end{cases} \]因此\[a_3=S_3-S_2=-\dfrac{1}{2^4}=-\dfrac{1}{16},\]而\[S_1+S_2+\cdots+S_{100}=\dfrac{-\dfrac 14-\left(-\dfrac{1}{2^{100}}\right)\cdot \dfrac14}{1-\dfrac14}=\dfrac 1 3 \left(\dfrac 1 {2^{100}} -1\right).\]