Tìm số nguyên dương n sao cho:

${\log _{2018}}2019 + {2^2}{\log _{\sqrt {2018} }}2019 + {3^2}{\log _{\sqrt[3]{{2018}}}}2019 + ... + {n^2}{\log _{\sqrt[n]{{2018}}}}2019$

= 1010² . 2021² log ₂₀₁₈ 2019

Lời giải

${\log _{2018}}2019 + {2^2}{\log _{\sqrt {2018} }}2019 + {3^2}{\log _{\sqrt[3]{{2018}}}}2019 + ... + {n^2}{\log _{\sqrt[n]{{2018}}}}2019$

$= {\log _{2018}}2019 + {2^2}\,.\,2{\log _{2018}}2019 + {3^2}\,.\,3{\log _{2018}}2019 + ... + {n^2}\,.\,n{\log _{2018}}2019$

= log ₂₀₁₈ 2019 + 2³ . log ₂₀₁₈ 2019 + 3³ . log ₂₀₁₈ 2019 + … + n³ . log ₂₀₁₈ 2019

= (1³ + 2³ + 3³ + … + n³) log ₂₀₁₈ 2019

Nên để ${\log _{2018}}2019 + {2^2}{\log _{\sqrt {2018} }}2019 + {3^2}{\log _{\sqrt[3]{{2018}}}}2019 + ... + {n^2}{\log _{\sqrt[n]{{2018}}}}2019$

= 1010² . 2021² log ₂₀₁₈ 2019 thì:

1³ + 2³ + 3³ + … + n³ = 1010² . 2021²

$\Rightarrow {\left( {\frac{{{n^2} + n}}{2}} \right)^2} = {1010^2}\,.\,{2021^2}$

$\Rightarrow \frac{{n\left( {n + 1} \right)}}{2} = 1010\,.\,2021$

Û n(n + 1) = 2 . 1010 . 2021 = 2020 . 2021

Þ n = 2020