Question

[GIA BẢO PHAN ĐỨC]

So sánh A=1/4^2+1/6^2+1/8^2+....+1/2024^2 với 1/4

Answer 1

[Akai Haruma]

Lời giải:

$A=\frac{1}{2^2}(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1012^2})$

$<\frac{1}{4}(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1011.1012})$

$=\frac{1}{4}(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1011}-\frac{1}{1012})$

$=\frac{1}{4}(1-\frac{1}{1012})$

$=\frac{1}{4}-\frac{1}{4.1012}< \frac{1}{4}$