Question

chung minh rang 1\4² +1\6²+1\8²+ .......+1\(2.n)²<1\4

Answer 1

ta có:

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

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

mà 1/2^2+1/3^2+..+1/n^2 < 1(cái này bn tự c/nm đc chứ?)

=>\(\frac{1}{4}.\left(\frac{1}{2^2}+\frac{1}{3^2}+..+\frac{1}{n^2}\right)<\frac{1}{4}\left(đpcm\right)\)

Answer 2

very sorry mik mới lớp 5 à nếu biết mik sẽ giải giùm bạn ! ^_^

Answer 3

chúc mừng năm mới nha