Cho n là số nguyên dương, tìm n sao cho:

\[{\log _a}2019 + {2^2}{\log _{\sqrt a }}2019 + {3^2}{\log _{\sqrt[3]{a}}}2019 + ... + {n^2}{\log _{\sqrt[n]{a}}}2019 = {1008^2}\,.\,{2017^2}{\log _a}2019\]

Lời giải

\[{\log _a}2019 + {2^2}{\log _{\sqrt a }}2019 + {3^2}{\log _{\sqrt[3]{a}}}2019 + ... + {n^2}{\log _{\sqrt[n]{a}}}2019\]

\( = {\log _a}2019 + {2^2}\,.\,2{\log _a}2019 + {3^2}\,.\,3{\log _a}2019 + ... + {n^2}\,.\,n{\log _a}2019\)

= log _a 2019 + 2³ . log _a 2019 + 3³ . log _a 2019 + … + n³ . log _a 2019

= (1³ + 2³ + 3³ + … + n³) log _a 2019

Suy ra \[{\log _a}2019 + {2^2}{\log _{\sqrt a }}2019 + {3^2}{\log _{\sqrt[3]{a}}}2019 + ... + {n^2}{\log _{\sqrt[n]{a}}}2019\]

\[ = {1008^2}\,.\,{2017^2}{\log _a}2019\]

Khi: 1³ + 2³ + 3³ + … + n³ = 1008² . 2017²

\( \Rightarrow {\left( {\frac{{{n^2} + n}}{2}} \right)^2} = {1008^2}\,.\,{2017^2}\)

\( \Rightarrow \frac{{n\left( {n + 1} \right)}}{2} = 1008\,.\,2017\)

Û n(n + 1) = 2 . 1008 . 2017 = 2016 . 2017

Þ n = 2016