Ta có :
\(A=\dfrac{1}{1.300}+\dfrac{1}{2.301}+\dfrac{1}{3.302}+..................+\dfrac{1}{101.400}\)
\(299A=\dfrac{299}{1.300}+\dfrac{299}{2.301}+\dfrac{299}{3.302}+..................+\dfrac{299}{101.400}\)
\(299A=1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+.................+\dfrac{1}{101}-\dfrac{1}{400}\)
\(299A=\left(1+\dfrac{1}{2}+.................+\dfrac{1}{101}\right)-\left(\dfrac{1}{300}+\dfrac{1}{301}+.............+\dfrac{1}{400}\right)=C\)
\(\Rightarrow A=\dfrac{C}{299}\)
Lại có :
\(B=\dfrac{1}{1.102}+\dfrac{1}{2.103}+................+\dfrac{1}{299.400}\)
\(101B=\dfrac{101}{1.102}+\dfrac{101}{2.103}+...............+\dfrac{101}{299.400}\)
\(101B=1-\dfrac{1}{102}+\dfrac{1}{2}-\dfrac{1}{103}+...............+\dfrac{1}{299}-\dfrac{1}{400}\)
\(101B=\left(1+\dfrac{1}{2}+...............+\dfrac{1}{299}\right)-\left(\dfrac{1}{102}+\dfrac{1}{103}+...............+\dfrac{1}{400}\right)=C\)\(\Rightarrow B=\dfrac{C}{101}\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{C}{101}:\dfrac{C}{299}=\dfrac{299}{101}\)
~ Chúc bn học tốt ~
