Đáp án:
$\dfrac{115}{33}$
Giải thích các bước giải:
$\dfrac{3}{1}+\dfrac{3}{3.5}+...+\dfrac{3}{97.99}\\
=3.\left (\dfrac{1}{1}+\dfrac{1}{3.5}+...+\dfrac{1}{97.99} \right )\\
=\dfrac{3}{2}.\left (\dfrac{2}{1}+\dfrac{2}{3.5}+...+\dfrac{2}{97.99} \right )\\
=\dfrac{3}{2}.\left ( 2+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99} \right )\\
=\dfrac{3}{2}.\left ( 2+\dfrac{1}{3}-\dfrac{1}{99} \right )\\
=\dfrac{3}{2}.\left ( \dfrac{2.99}{99}+\dfrac{33}{99}-\dfrac{1}{99} \right )\\
=\dfrac{3}{2}.\dfrac{230}{99}\\
=\dfrac{115}{33}$
