Ta có : \(\dfrac{1}{2^2}=\dfrac{1}{2\times2}< \dfrac{1}{1\times2}\\ \dfrac{1}{3^2}=\dfrac{1}{3\times3}< \dfrac{1}{2\times3}\\ \dfrac{1}{4^2}=\dfrac{1}{4\times4}< \dfrac{1}{3\times4}\\ ...\\ \dfrac{1}{100^2}=\dfrac{1}{100\times100}< \dfrac{1}{99\times100}\)
\(\Rightarrow\)\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{99\times100}\)
hay \(A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{100}\)
\(\Rightarrow A< \dfrac{100}{100}-\dfrac{1}{100}\)
\(\Rightarrow A< \dfrac{99}{100}\)
Mà \(\dfrac{99}{100}< 1\)
\(\Rightarrow A< 1\)
Vậy \(A< 1\)(đpcm)
Ta có : \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
...............
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}< 1\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1\)
Vậy A<1
ta có : A< \(\dfrac{1}{1\cdot2}\) +\(\dfrac{1}{2\cdot3}\) +......+\(\dfrac{1}{99\cdot100}\)
A< 1-\(\dfrac{1}{2}\) +\(\dfrac{1}{2}\) -\(\dfrac{1}{3}\) +......+\(\dfrac{1}{99}\) -\(\dfrac{1}{100}\)
A<1-\(\dfrac{1}{100}\)
A< \(\dfrac{99}{100}\) < 1
\(\Rightarrow\) A<1
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2\cdot2}< \dfrac{1}{1\cdot2}\\ \dfrac{1}{3^2}=\dfrac{1}{3\cdot3}< \dfrac{1}{2\cdot3}\\ \dfrac{1}{4^2}=\dfrac{1}{4\cdot4}< \dfrac{1}{3\cdot4}\\ .........\\ \dfrac{1}{100^2}=\dfrac{1}{100\cdot100}< \dfrac{1}{99\cdot100}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ \Rightarrow A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ \Rightarrow A=\dfrac{1}{1}-\dfrac{1}{100}\\ \Rightarrow A=\dfrac{99}{100}\\ \Rightarrow A< \dfrac{100}{100}hayA< 1\left(đpcm\right)\)
Hoang Hung Quan Nguyễn Huy Tú soyeon_Tiểubàng giải Tuấn Anh Phan Nguyễn
