Toán Chứng Minh rằng 1/3+1/3^2+….+1/3^2019+1/3^2020 < 1/2 09/09/2021 By Amara Chứng Minh rằng 1/3+1/3^2+….+1/3^2019+1/3^2020 < 1/2
\[\begin{array}{l} A = \frac{1}{3} + \frac{1}{{{3^2}}} + … + \frac{1}{{{3^{2019}}}} + \frac{1}{{{3^{2020}}}}\\ \Rightarrow 3A = 1 + \frac{1}{3} + \frac{1}{{{3^2}}} + … + \frac{1}{{{3^{2019}}}}\\ \Rightarrow 3A – A = 1 – \frac{1}{{{3^{2020}}}}\\ \Rightarrow 2A = 1 – \frac{1}{{{3^{2020}}}}\\ \Rightarrow A = \frac{1}{2} – \frac{1}{2}.\frac{1}{{{3^{2020}}}}\\ \frac{1}{2}.\frac{1}{{{3^{2020}}}} > 0 \Rightarrow A < \frac{1}{2}\,\,\left( {dpcm} \right) \end{array}\] Trả lời
$A=\frac{1}{3}+$ $\frac{1}{3^2}+…+$ $\frac{1}{3^{2020}}$ $⇒3A=1+\frac{1}{3}+…+$ $\frac{1}{3^{2019}}$ $⇒3A-A=1+(\frac{1}{3}-$ $\frac{1}{3})+…-$ $\frac{1}{3^{2020}}$ $⇒2A=1-\frac{1}{3^{2020}}$ $⇒A=\frac{1}{2}-$ $\frac{1}{2}.$ $\frac{1}{3^{2020}}<$ $\frac{1}{2}$ Trả lời
\[\begin{array}{l}
A = \frac{1}{3} + \frac{1}{{{3^2}}} + … + \frac{1}{{{3^{2019}}}} + \frac{1}{{{3^{2020}}}}\\
\Rightarrow 3A = 1 + \frac{1}{3} + \frac{1}{{{3^2}}} + … + \frac{1}{{{3^{2019}}}}\\
\Rightarrow 3A – A = 1 – \frac{1}{{{3^{2020}}}}\\
\Rightarrow 2A = 1 – \frac{1}{{{3^{2020}}}}\\
\Rightarrow A = \frac{1}{2} – \frac{1}{2}.\frac{1}{{{3^{2020}}}}\\
\frac{1}{2}.\frac{1}{{{3^{2020}}}} > 0 \Rightarrow A < \frac{1}{2}\,\,\left( {dpcm} \right) \end{array}\]
$A=\frac{1}{3}+$ $\frac{1}{3^2}+…+$ $\frac{1}{3^{2020}}$
$⇒3A=1+\frac{1}{3}+…+$ $\frac{1}{3^{2019}}$
$⇒3A-A=1+(\frac{1}{3}-$ $\frac{1}{3})+…-$ $\frac{1}{3^{2020}}$
$⇒2A=1-\frac{1}{3^{2020}}$
$⇒A=\frac{1}{2}-$ $\frac{1}{2}.$ $\frac{1}{3^{2020}}<$ $\frac{1}{2}$