高中数学选择性必修 第二册（381题）

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知识点：第四章 数列

参考答案：解：\(\because {b_n} = \frac{4}{{{a_n} \cdot {a_{n + 1}}}} = \frac{4}{{\left( {4n - 2} \right)\left( {4n + 2} \right)}} = \frac{1}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}} = \frac{1}{2}\left( {\frac{1}{{2n - 1}} - \frac{1}{{2n + 1}}} \right)\)，\(\therefore {S_n} = \frac{1}{2}\left[ {\left( {1 - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{5}} \right) + \left( {\frac{1}{5} - \frac{1}{7}} \right) + \cdots + \left( {\frac{1}{{2n - 1}} - \frac{1}{{2n + 1}}} \right)} \right] = \frac{1}{2}\left( {1 - \frac{1}{{2n + 1}}} \right) = \frac{n}{{2n + 1}}\).