Ta có: a=1/1.300+1/2.301+...+1/101.400
⇒ a= 1/299.(299/1.300+299/2.301+...+299/101.400)
⇒ a= 1/299. ( 1+1/300+1/2-1/301+....+1/101-1/400)
⇒ a= 1/299.|(1+1/2+....+1/101)-(1/300+1/301+....+1/400)|
Ta có: b=1/1.102+1/2.103+..+1/299.400
⇒ b= 1/101.(101/1.102+101/2.103+..+101/299.400)
⇒ 1/101.|(1-1/102+1/2-1/102+......+1/299-1/400)|
⇒ b= 1/101 .|(1+1/2+....+1/299) - (1/102+1/103+....+1/400)|
⇒ b= |(1+1/2+....+1/299)- (1/300+1/301+....+1/400)|
⇒a=1/299.|(1+1/2+....+1/101)-(1/300+1/301+....+1/400)|
phần
b=1/101.|(1+1/2+....+1/101)-(1/300+1/301+....+1/400)|
⇒a/b=1/299:1/101
⇒a/b=101/299.
Ta chú ý đẳng thức \(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\)(Chứng minh rất dễ, bạn quy đồng lên là được nha)
\(A=\frac{1}{1.300}+\frac{1}{2.301}+...+\frac{1}{101.400}\)
\(\Rightarrow299A=\frac{299}{1.300}+\frac{299}{2.301}+\frac{299}{3.302}+...+\frac{299}{101+400}\)
\(=1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+\frac{1}{3}-\frac{1}{302}+...+\frac{1}{101}-\frac{1}{400}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\)
Đặt \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}=X,\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}=Y\)
\(\Rightarrow A=\frac{X-Y}{299}\)
\(B=\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}\)
\(\Rightarrow101B=\frac{101}{1.102}+\frac{101}{2.103}+\frac{101}{3.104}+...+\frac{101}{299.400}\)
\(=1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+\frac{1}{3}-\frac{1}{104}+...+\frac{1}{102}-\frac{1}{203}+\frac{1}{103}-\frac{1}{204}+...\)
\(\frac{1}{198}-\frac{1}{299}+\frac{1}{199}-\frac{1}{300}+\frac{1}{200}-\frac{1}{301}+...+\frac{1}{299}-\frac{1}{400}\)
\(=\left(1+...+\frac{1}{101}\right)-\left(\frac{1}{300}+...+\frac{1}{400}\right)+\left(\frac{1}{102}-\frac{1}{102}\right)+\left(\frac{1}{103}-\frac{1}{103}\right)+...+\left(\frac{1}{299}-\frac{1}{299}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)=X-Y\)
\(\Rightarrow B=\frac{X-Y}{101}>\frac{X-Y}{299}=A\)
Vậy \(B>A\)
