cau hoi tuong tu day ban nhe

Bản đẹp :

CMR : \(\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+.....+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{4}\)

Ta có: 

N = \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}=\frac{1}{\left(2.2\right)^2}+\frac{1}{\left(2.3\right)^2}+\frac{1}{\left(2.4\right)^2}+...+\frac{1}{\left(2n\right)^2}\)

= \(\frac{1}{2^2.2^2}+\frac{1}{2^2.3^2}+\frac{1}{2^2.4^2}+...+\frac{1}{2^2.n^2}=\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)\)

Mà \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}<1\) (lát nữa mình sẽ chứng minh)

=> N <\(\frac{1}{4}.1=\frac{1}{4}\)

Ta sẽ chứng minh bổ đề: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}<1\)

Thật vậy:

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)= \(1-\frac{1}{n^2}<1\)

A=1/4^2+1/6^2+...+1/(2n)^2

=1/4(1/2^2+1/3^2+...+1/n^2)

=>A<1/4(1-1/2+1/2-1/3+...+1/n-1-1/n)

=>A<1/4(1-1/n)<1/4

Ta có:

\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}\)

\(=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{\left(2n-2\right).2n}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{12}+...+\frac{1}{2n-2}-\frac{1}{2n}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2n}\right)\)

\(=\frac{1}{4}-\frac{1}{2n.2}\)

Vì \(\frac{1}{4}-\frac{1}{2n.2}< \frac{1}{4}\)

\(\Rightarrow\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}< \frac{1}{4}\)

Vậy \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}< \frac{1}{4}\) (Đpcm)

BƯỚC 1: THẤY ĐÚNG ĐÂU:

(....)=(...) KHÔNG HIỂU . bước này nhằm mục đích gì