chứng minh N=1/4 mũ 2 +1/6 mũ 2 +1/8 mũ 2 +……..+1/2n mũ 2 >1/4

Đáp án:

$N=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+…+\frac{1}{(2n)^2}\\
=\frac{1}{2^2}\left (\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+…+\frac{1}{n^2}  \right )$
Mà $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+…+\frac{1}{n^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+…+\frac{1}{(n-1).n}\\
= 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+…+\frac{1}{n-1}-\frac{1}{n}\\
=1-\frac{1}{n}<1\\
\Rightarrow \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+…+\frac{1}{n^2}<1\\
\Rightarrow N<\frac{1}{2^2}.1=\frac{1}{4}$