Question

[Hoàng Bảo]

Cho $A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\mathrm{...}+\frac{1}{{100}^2}$. Chứng minh $A<\frac{1}{2}$.

Answer 1

[Akai Haruma]

Lời giải:

$A=\frac{1}{2^2}(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2})$

$=\frac{1}{4}(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2})$

$<\frac{1}{4}(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50})$
$=\frac{1}{4}(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50})$

$=\frac{1}{4}(1+1-\frac{1}{50})=\frac{1}{4}(2-\frac{1}{50})< \frac{1}{4}.2=\frac{1}{2}$
Ta có đpcm.