共50条信息
\(1-\dfrac{1}{2}=\dfrac{1}{2}\);\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{1}{3}+\dfrac{1}{4}\);\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}\);\(\cdots \quad \quad \cdots \)据此规律,第\(n\)个等式可为:_________________________________.
数列\(1\dfrac{1}{2},3\dfrac{1}{4},5\dfrac{1}{8},7\dfrac{1}{16},\cdots \)的前\(n\)项和\({{S}_{n}}\)为
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