Đáp án:
Giải thích các bước giải:
$\quad\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{45}+...+\dfrac{1}{(2n-1)(2n)}\\
=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n}\\
=1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{2n-1}-(\dfrac{1}{2}+\dfrac{1}{4}+..+\dfrac{1}{2n})\\
=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2n-1}+\dfrac{1}{2n}-2(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{2n})\\
=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2n-1}+\dfrac{1}{2n}-(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n})\\
=\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{2n}$
