Chứng Minh rằng 1/3+1/3^2+....+1/3^2019+1/3^2020 < 1/2
2 câu trả lời
\[\begin{array}{l} A = \frac{1}{3} + \frac{1}{{{3^2}}} + ... + \frac{1}{{{3^{2019}}}} + \frac{1}{{{3^{2020}}}}\\ \Rightarrow 3A = 1 + \frac{1}{3} + \frac{1}{{{3^2}}} + ... + \frac{1}{{{3^{2019}}}}\\ \Rightarrow 3A - A = 1 - \frac{1}{{{3^{2020}}}}\\ \Rightarrow 2A = 1 - \frac{1}{{{3^{2020}}}}\\ \Rightarrow A = \frac{1}{2} - \frac{1}{2}.\frac{1}{{{3^{2020}}}}\\ \frac{1}{2}.\frac{1}{{{3^{2020}}}} > 0 \Rightarrow A < \frac{1}{2}\,\,\left( {dpcm} \right) \end{array}\]
$A=\frac{1}{3}+$ $\frac{1}{3^2}+...+$ $\frac{1}{3^{2020}}$
$⇒3A=1+\frac{1}{3}+...+$ $\frac{1}{3^{2019}}$
$⇒3A-A=1+(\frac{1}{3}-$ $\frac{1}{3})+...-$ $\frac{1}{3^{2020}}$
$⇒2A=1-\frac{1}{3^{2020}}$
$⇒A=\frac{1}{2}-$ $\frac{1}{2}.$ $\frac{1}{3^{2020}}<$ $\frac{1}{2}$
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