Question

F=\(\dfrac{1}{1x4}\)+\(\dfrac{1}{4x7}\)+....+\(\dfrac{1}{97x100}\)+\(\dfrac{1}{100x103}\)

Answer 1

Giờ anh đang bận hồi nữa anh giúp cho nha

Answer 2

\(F=\dfrac{1}{x}\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}+\dfrac{1}{100.103}\right)\)

\(3F=\dfrac{1}{x}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}+\dfrac{3}{100.103}\right)\)

\(F=\dfrac{\dfrac{1}{x}\left(\dfrac{1}{3}-\dfrac{1}{103}\right)}{3}=\dfrac{\dfrac{1}{x}.\dfrac{100}{309}}{3}=\dfrac{\dfrac{100x}{309}}{3}=\dfrac{100x}{927}\)

Answer 3

\(F=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{103}\right)=\dfrac{1}{3}.\dfrac{102}{103}=\dfrac{204}{309}\)