Đáp án:
Giải thích các bước giải:
Ta có:
$A=\cfrac{1}{5}+\cfrac{2}{5^2}+\cfrac{3}{5^3}+...\cfrac{2020}{5^{2020}}\\\Leftrightarrow \cfrac{1}{5}A=\cfrac{1}{5^2}+\cfrac{2}{5^3}+\cfrac{3}{5^4}+...+\cfrac{2020}{5^{2021}}\\\Rightarrow A-\cfrac{1}{5}A=\cfrac{1}{5}+\cfrac{1}{5^2}+\cfrac{1}{5^3}+...+\cfrac{1}{5^{2020}}-\cfrac{2020}{5^{2021}}=\cfrac{4}{5}A$
Xét tổng:
$B=\cfrac{1}{5}+\cfrac{1}{5^2}+\cfrac{1}{5^3}+...+\cfrac{1}{5^{2020}}\\\Leftrightarrow \cfrac{1}{5}B=\cfrac{1}{5^2}+\cfrac{1}{5^3}+\cfrac{1}{5^4}+...+\cfrac{1}{5^{2021}}\\\Rightarrow \cfrac{1}{5}B-B=\cfrac{1}{5^{2021}}-\cfrac{1}{5}=-\cfrac{4}{5}B\\\Leftrightarrow B=\cfrac{-5}{4\times 5^{2021}}+\cfrac{1}{4}$
Suy ra:
$\cfrac{4}{5}A=\cfrac{-5}{4\times 5^{2021}}+\cfrac{1}{4}-\cfrac{2020}{5^{2021}}\\=\cfrac{-8085}{4\times 5^{2021}}+\cfrac{1}{4}\\\Leftrightarrow A=\cfrac{5}{16}-\cfrac{8085}{16\times 5^{2020}}$
Do $-\cfrac{8085}{16\times 5^{2020}}<0$ và $\cfrac{5}{16}<\cfrac{1}{3}$ nên:
$\cfrac{5}{16}-\cfrac{8085}{16\times 5^{2020}}<\cfrac{1}{3}\\\Leftrightarrow A<\cfrac{1}{3}$
Ta có:
$1617<5^{1617}\\\Rightarrow 1617<5^{2019}\\\Rightarrow 8085<5^{2020}\\\Rightarrow \cfrac{8085}{5^{2020}}<1\\\Rightarrow \cfrac{8085}{16\times 5^{2020}}<\cfrac{1}{16}\\\Rightarrow -\cfrac{8085}{16\times 5^{2020}}>-\cfrac{1}{16}\\\Rightarrow \cfrac{5}{16}-\cfrac{8085}{16\times 5^{2020}}>-\cfrac{1}{16}+\cfrac{5}{16}=\cfrac{1}{4}\\\Rightarrow A>\cfrac{1}{4}$
Suy ra: $\cfrac{1}{4}<\cfrac{1}{5}+\cfrac{2}{5^2}+\cfrac{3}{5^2}+...+\cfrac{2020}{5^{2020}}<\cfrac{1}{3}$
