Đáp án:
\(Q<\frac{1}{2}(ĐPCM\)
Giải thích các bước giải:
Q=\(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+...+\frac{1}{3^{2019}}+\frac{1}{3^{2020}}\)
⇒\(2Q=3(1+\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+...+\frac{1}{3^{2019}}+\frac{1}{3^{2020}})\)
⇒\(2Q=1+\frac{1}{3}+\frac{1}{3^{2}}+...+\frac{1}{3^{2019}}+\frac{1}{3^{2020}}\)
⇒\(2Q-Q=(1+\frac{1}{3}+...+\frac{1}{3^{2019}})-(\frac{1}{3}+\frac{1}{3^{2}}+...+\frac{1}{3^{2020}}\)
⇒\(Q=1-\frac{1}{3^{2020}}<\frac{1}{2}\)
⇒\(Q<\frac{1}{2}(ĐPCM\)
