Question

Chứng minh rằng \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}<1\)

Answer 1

chưa học nên chưa biết.-------------

Answer 2

Ta co \(\frac{1}{2^2}<\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2};\frac{1}{3^2}<\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3};...;\frac{1}{100^2}<\frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}<\frac{1}{1}-\frac{1}{100}<\frac{1}{1}=1\)

\(\Rightarrow dpcm\)

Tick cho to nhe Nguyen Huu Cuong

Ma cau thi hsg a? bai nay danh cho hoc sinh gioi ma

Answer 3

bai thi .....................kho..........................kho..............troi.................thilanh.............................ret..................wa.........................dau................wa......................tich....................ung.....................ho.....................cho............do.................lanh