已知数列 \(\{ {a_n}\} \) 的前 \(n\) 项和为 \({S_n}\) ， \({a_1}=1\) ，且 \({S_n}=\lambda {a_n} - 1\) （λ为常数）．若数列 \(\{ {b_n}\} \) 满足 \({a_n}{b_n}= - {n^2} + 9n - 20\) ，且 \({b_{n + 1}} < {b_n}\) ，则满足条件的\(n\)的取值可以为（       ）

当 \(n = 1\) 时， \({a_1} = {S_1} = \lambda {a_1} - 1\) ， \(\therefore \lambda - 1 = 1\) ，解得： \(\lambda = 2\)

\(\therefore {S_n} = 2{a_n} - 1\) 

当 \(n \geqslant 2\) 且 \(n \in {N^ * }\) 时， \({S_{n - 1}} = 2{a_{n - 1}} - 1\) 

\(\therefore {a_n} = {S_n} - {S_{n - 1}} = 2{a_n} - 2{a_{n - 1}}\) ，即： \({a_n} = 2{a_{n - 1}}\) 

\(\therefore \)数列 \(\left\{ {{a_n}} \right\}\) 是以\(1\)为首项， \(2\) 为公比的等比数列， \(\therefore {a_n} = {2^{n - 1}}\) 

\(\because {a_n}{b_n} = - {n^2} + 9n - 20\) ， \(\therefore {b_n} = \frac{{ - {n^2} + 9n - 20}}{{{2^{n - 1}}}}\) 

\(\therefore {b_{n + 1}} - {b_n} = \frac{{ - {{\left( {n + 1} \right)}^2} + 9\left( {n + 1} \right) - 20}}{{{2^n}}} - \frac{{ - {n^2} + 9n - 20}}{{{2^{n - 1}}}} = \frac{{{n^2} - 11n + 28}}{{{2^n}}} < 0\) 

\(\because {2^n} > 0\) ， \(\therefore {n^2} - 11n + 28 = \left( {n - 4} \right)\left( {n - 7} \right) < 0\) ，解得： \(4 < n < 7\) 

又 \(n \in {N^ * }\) ， \(\therefore n = 5\)或\(6\) 

故选：AB

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