Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

...

\(\frac{1}{2022^2}<\frac{1}{2021\cdot2022}=\frac{1}{2021}-\frac{1}{2022}\)

Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2022^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{2021}-\frac{1}{2022}\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2022^2}<1-\frac{1}{2022}<1\) (ĐPCM)

Sửa đề: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{100^2}<\frac34\)

Ta có: \(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

\(\frac{1}{4^2}<\frac{1}{3\cdot4}=\frac13-\frac14\)

...

\(\frac{1}{100^2}<\frac{1}{99\cdot100}=\frac{1}{99}-\frac{1}{100}\)

Do đó: \(\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{100^2}<\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{99}-\frac{1}{100}=\frac12-\frac{1}{100}<\frac12\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{100^2}<\frac14+\frac12=\frac34\)

bạn ơi qua giúp mk với

mk viết nhầm 

A = 1 / 2² + 1 / 3² + 1 / 4² + ... + 1 / 80² mới đúng nhé

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)