Cho dãy số (un) có số hạng đầu \({u_1} \ne 1\) và thỏa mãn \(\log _2^2\left( {5{u_1}} \right) + \log _2^2\left( {7{u_1}} \right) = \log _2^25 + \log _2^27\). Biết \({u_{n + 1}} = 7...

* Hướng dẫn giải

\(\log _2^2\left( {5{u_1}} \right) + \log _2^2\left( {7{u_1}} \right) = \log _2^25 + \log _2^27 \Leftrightarrow \log _2^2\left( {5{u_1}} \right) - \log _2^25 + \log _2^2\left( {7{u_1}} \right) - \log _2^27 = 0\)

\( \Leftrightarrow \left( {{{\log }_2}5{u_1} - {{\log }_2}5} \right)\left( {lo{g_2}5{u_1} + {{\log }_2}5} \right) + \left( {{{\log }_2}7{u_1} - {{\log }_2}7} \right)\left( {{{\log }_2}7{u_1} + {{\log }_2}7} \right) = 0\)

\( \Leftrightarrow {\log _2}\left( {{u_1}} \right).{\log _2}\left( {25{u_1}} \right) + {\log _2}\left( {{u_1}} \right).{\log _2}\left( {49{u_1}} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} {\log _2}{u_1} = 0\\ {\log _2}\left( {25{u_1}} \right) + {\log _2}\left( {49{u_1}} \right) = 0 \end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l} {u_1} = 1\\ {\log _2}\left( {1225u_1^2} \right) = 0 \end{array} \right. \Leftrightarrow 1225u_1^2 = 1 \Leftrightarrow u_1^2 = \frac{1}{{1225}} \Rightarrow {u_1} = \frac{1}{{35}}\)

Lại có \({u_{n + 1}} = 7{u_n} \Rightarrow \left( {{u_n}} \right)\) là cấp số nhân với \({u_1} = \frac{1}{{35}};q = 7 \Rightarrow {u_n} = \frac{{{7^{n - 1}}}}{{35}}\)

Do đó \({u_n} > 1111111 \Leftrightarrow \frac{{{7^{n - 1}}}}{{35}} > 1111111 \Leftrightarrow n > 1 + {\log _7}\left( {35.1111111} \right) \approx 9,98.\)