Tổng \(\begin{aligned}
&S=1+2^{2} \log _{\sqrt{2}} 2+3^{2} \log _{\sqrt[3]{2}} 2+\ldots .+2018^{2} \log _{2019 \sqrt{2}} 2=1+2^{2} \log _{2^{\frac{1}{2}}} 2+3^{2} \log _{2^{\frac{1}{3}}} 2+\ldots .+2018^{2} \log _{2^{\frac{1}{2018}}} 2
\end{aligned}\)
là:
Suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Lời giải:
Báo sai\(\begin{aligned} &\text { Ta có } 1^{3}+2^{3}+3^{3}+\ldots+n^{3}=\frac{(n(n+1))^{2}}{4} \text { . }\\ &\text { Mặt khác }\\ &S=1+2^{2} \log _{\sqrt{2}} 2+3^{2} \log _{\sqrt[3]{2}} 2+\ldots .+2018^{2} \log _{2019 \sqrt{2}} 2=1+2^{2} \log _{2^{\frac{1}{2}}} 2+3^{2} \log _{2^{\frac{1}{3}}} 2+\ldots .+2018^{2} \log _{2^{\frac{1}{2018}}} 2\\ &=1+2^{3} \log _{2} 2+3^{3} \log _{2} 2+\ldots .+2018^{3} \log _{2} 2=1+2^{3}+3^{3}+\ldots+2018^{3}=\left[\frac{2018(2018+1)}{2}\right]^{2}\\ &=1009^{2} \cdot 2019^{2} \end{aligned}\)