Question

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

Answer 1

Ta có:

\(\frac{1}{2^2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}\)

...

\(\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(\Leftrightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Leftrightarrow A< 1-\frac{1}{100}< 1\left(đpcm\right)\)

Answer 2

giúp mk vs các bạn ưi ! mk đang cần gấp ai nhanh mik tích cho !nhanh nha help me!thank nhìu