Đáp án: $A=\dfrac{337}{1010}$
Giải thích các bước giải:
Ta có:
$I_n=1-\dfrac{1}{1+2+3+...+n}$
$\to I_n=\dfrac{1+2+3+...+n-1}{1+2+3+...+n}$
$\to I_n=\dfrac{2+3+...+n}{1+2+3+...+n}$
$\to I_n=\dfrac{\dfrac{\left(n-1\right)\left(n+2\right)}{2}}{\dfrac{n\left(n+1\right)}{2}}$
$\to I_n=\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}$
Khi đó
$A=\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)....\left(1-\dfrac{1}{1+2+3+...+2020}\right)$
$\to A=\dfrac{\left(2-1\right)\left(2+2\right)}{2\left(2+1\right)}\cdot \dfrac{\left(3-1\right)\left(3+2\right)}{3\left(3+1\right)}....\dfrac{\left(2020-1\right)\left(2020+2\right)}{2020\left(2020+1\right)}$
$\to A=\dfrac{1\cdot 4}{2\cdot 3}\cdot \dfrac{2\cdot 5}{3\cdot 4}....\dfrac{2019\cdot 2022}{2020\cdot 2021}$
$\to A=\dfrac{1.2...2019}{2.3...2020}\cdot \dfrac{4.5...2022}{3.4...2021}$
$\to A=\dfrac{1}{2020}\cdot \dfrac{2022}{3}$
$\to A=\dfrac{337}{1010}$
