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