Question

[0o0_Cô Nàng Năng Động_0o0]

Cho A=\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...........+\frac{1}{1+2+3+\text{...+99}}\)

B=\(1+\frac{1}{2^2}+\frac{1}{3^2}+.......+\frac{1}{49^2}\)

Chứng minh A<B

Answer 1

[Phùng Minh Quân]

Ta có : 

\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+3}+...+\frac{1}{1+2+3+...+99}\)

\(A=\frac{1}{\frac{2\left(2+1\right)}{2}}+\frac{1}{\frac{3\left(3+1\right)}{2}}+\frac{1}{\frac{4\left(4+1\right)}{2}}+...+\frac{1}{\frac{99\left(99+1\right)}{2}}\)

\(A=\frac{2}{2\left(2+1\right)}+\frac{2}{3\left(3+1\right)}+\frac{2}{4\left(4+1\right)}+...+\frac{2}{99\left(99+1\right)}\)

\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{99.100}\)

\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(A=2.\frac{49}{100}\)

\(A=\frac{49}{50}\)

Lại có : 

\(\frac{1}{2^2}>\frac{1}{2.3}\)

\(\frac{1}{3^2}>\frac{1}{3.4}\)

\(\frac{1}{4^2}>\frac{1}{4.5}\)

\(............\)

\(\frac{1}{49^2}>\frac{1}{49.50}\)

\(\Rightarrow\)\(B=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{49^2}>1+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)

\(B>1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)

\(B>1+\frac{1}{2}-\frac{1}{50}\)

\(B>1+\frac{12}{25}=\frac{37}{25}=\frac{74}{50}>\frac{49}{50}=A\)

\(\Rightarrow\)\(B>A\)

Vậy \(A< B\)

Chúc bạn học tốt ~