已知数列 \( \left\{{a}_{n}\right\}\) 是等比数列， \({a_2} = 2\) ， \( {a}_{5}=\frac{1}{4}\) ，则\( {a}_{1}\cdot {a}_{2}\cdot {a}_{3}+{a}_{2}\cdot {a}_{3}\cdot {a}_{4}+\cdot \cdot \cdot +{a}_{n}\cdot {a}_{n+1}\cdot {a}_{n+2}=\) ___.

设数列 \( \left\{{a}_{n}\right\}\) 的公比为 \( q\) ，则 \({q^3} = \frac{{{a_5}}}{{{a_2}}} = \frac{1}{8}\) ，解得 \( q=\frac{1}{2}\) ，

∴ \({a_1} = 4\) ， \( {a}_{3}=1\) ，∴ \( {a}_{1}\cdot {a}_{2}\cdot {a}_{3}=4\times 2\times 1=8\) ，

\( \frac{{a}_{n+1}{a}_{n+2}{a}_{n+3}}{{a}_{n}{a}_{n+1}{a}_{n+2}}=\frac{{a}_{n}q·{a}_{n+1}q·{a}_{n+2}q}{{a}_{n}{a}_{n+1}{a}_{n+2}}={q}^{3}\) ,

∴数列 \( \left\{{a}_{n}{a}_{n+1}{a}_{n+2}\right\}\) 是首项为 \({a_1} \cdot {a_2} \cdot {a_3} = 8\) ，公比为 \( {q}^{3}=\frac{1}{8}\) 的等比数列，

∴ \( {a}_{1}\cdot {a}_{2}\cdot {a}_{3}+{a}_{2}\cdot {a}_{3}\cdot {a}_{4}+\cdot \cdot \cdot +{a}_{n}\cdot {a}_{n+1}\cdot {a}_{n+2}=\frac{8\times (1-\frac{1}{{8}^{n}})}{1-\frac{1}{8}}=\frac{64}{7}(1-{2}^{-3n})\) .

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