Question

[tran thi phuong thao]

chúng minh rằng :

1/2² + 1/3² + 1/4² + ... + 1/100²   < 1

Giúp mik nha , thanks!

Answer 1

[Trần Thanh Phương]

Dễ thấy \(\frac{1}{2^2}< \frac{1}{1\cdot2};...;\frac{1}{100^2}< \frac{1}{99\cdot100}\)

Do đó : \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}< 1\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)

Answer 2

[hoc hieu qua]

Ta có : \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}221​+321​+421​+...+10021​<1.21​+2.31​+3.41​+...+99.1001​
        =1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}< 1=1−21​+21​−31​+31​−41​+...+991​−1001​=1−1001​<1