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高中数学选择性必修 第二册(381题)


已知 \(\left\{ {{a_n}} \right\}\) 为等比数列,前\(n\)项和为 \({S_n}\left( {n \in {N^*}} \right)\)\({a_1} = 1\)\({a_3}{a_4} = \frac{1}{{32}}\).



求 \(\left\{ {{a_n}} \right\}\) 的通项公式及前\(n\)项和 \({S_n}\)




知识点:第四章 数列


参考答案:解:设公比为\(q\),∵\({a_1} = 1\),\({a_3}{a_4} = \frac{1}{{32}}\),∴\({q^5} = \frac{1}{{32}}\),∴\(q = \frac{1}{2}\),∴\({a_n} = {\left( {\frac{1}{2}} \right)^{n - 1}}\);\({S_n} = \frac{{1 - {{\left( {\frac{1}{2}} \right)}^n}}}{{1 - \frac{1}{2}}} = 2 - {\left( {\frac{1}{2}} \right)^{n - 1}}\)

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