Đáp án: $S=\dfrac12-\dfrac{1}{4046}$
Giải thích các bước giải:
Ta có:
$S=\dfrac{1}{1^2+2.1}+\dfrac{1}{3^2+2.3}+...+\dfrac{1}{2021^2+2.2021}$
$\to S=\dfrac{1}{1(1+2)}+\dfrac{1}{3(3+2)}+...+\dfrac1{2021(2021+2)}$
$\to S=\dfrac1{1.3}+\dfrac1{3.5}+...+\dfrac1{2021.2023}$
$\to 2S=\dfrac2{1.3}+\dfrac2{3.5}+...+\dfrac2{2021.2023}$
$\to 2S=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{2023-2021}{2021.2023}$
$\to 2S=\dfrac11-\dfrac13+\dfrac13-\dfrac15+...+\dfrac1{2021}-\dfrac{1}{2023}$
$\to 2S=1-\dfrac1{2023}$
$\to S=\dfrac12-\dfrac{1}{4046}$
