Đáp án:
Giải thích các bước giải:
$b)=\dfrac{1}{2}. (\dfrac{4}{3. 7}+ \dfrac{4}{7. 11}+ \dfrac{4}{11. 15}+...+ \dfrac{4}{47. 15}$
$= \dfrac{1}{2}. (\dfrac{1}{3}- \dfrac{1}{7}+ \dfrac{1}{7}- \dfrac{1}{11}+ \dfrac{1}{11}- \dfrac{1}{15}+...+ \dfrac{1}{47}- \dfrac{1}{51})$
$= \dfrac{1}{2}. (\dfrac{1}{3}. \dfrac{1}{51})$
$= \dfrac{1}{2}.\dfrac{16}{51}$
$= \dfrac{8}{51}$
