Question

[Huỳnh Thiện Tâm]

. Chứng tỏ rằng: 1 2 + 1 2 2 + 1 2 3 + 1 2 4 + ⋯ + 1 2 100 < 1

Answer 1

[Gái Việt đó]

Đề bài sai!

Answer 2

[Nguyễn Linh Nhi]

a có : $\frac{1}{2^2}<\frac{1}{1.2}$

$\frac{1}{3^2}<\frac{1}{2.3}$

$\frac{1}{4^2}<\frac{1}{3.4}$
...
$\frac{1}{{100}^2}<\frac{1}{99.100}$

=> $\frac{1}{2^2}$ + $\frac{1}{3^2}$ + ... + $\frac{1}{{100}^2}$ < $\frac{1}{1.2}$ + $\frac{1}{2.3}$ + ...+ $\frac{1}{99.100}$
Ta có : $\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}$

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

= 1 - $\frac{1}{100}=\frac{99}{100}<1$

=> $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{{100}^2}<1$