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高中数学选择性必修 第二册(381题)


已知无穷数列满足 \({a_{n + 1}} = p \cdot {a_n} + \frac{q}{{{a_n}}}\left( {n \in {{\rm{N}}^*}} \right)\) ,其中\(p,q\) 均为非负实数且不同时为0.


若 \(p = \frac{1}{2}\) , \(q = 2\) ,且 \({a_3} = \frac{{41}}{{20}}\) ,求 \({a_1}\) 的值;



知识点:第四章 数列


参考答案:∵ \(p = \frac{1}{2}\) , \(q = 2\) ,
∴ \({a_{n + 1}} = \frac{1}{2} \cdot {a_n} + \frac{2}{{{a_n}}}\) ,
\(a_n^2 - 2{a_n}{a_{n + 1}} + 4 = 0\) ,∵ \({a_3} = \frac{{41}}{{20}}\) ,
∴ \(a_2^2 - \frac{{41}}{{10}}{a_2} + 4 = 0\) ,
解得 \({a_2} = \frac{5}{2}\) 或 \({a_2} = \frac{8}{5}\) ,
当 \({a_2} = \frac{5}{2}\) 时,
\(a_1^2 - 5{a_1} + 4 = 0\) ,
解得 \({a_1} = 1\) 或 \({a_1} = 4\) ;
\({a_2} = \frac{8}{5}\) 时,
\(a_1^2 - \frac{{16}}{5}{a_1} + 4 = 0\) ,
无解.∴ \({a_1} = 1\) 或 \({a_1} = 4\) ;

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