Question

CHỨNG MINH RẰNG:

  \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+...+\frac{1}{\left(2n^2\right)}< \frac{1}{4}\)

Answer 1

bài này tui bít làm nhưng dài lắm

Answer 2

Cố gắng làm hộ mình với !

Answer 3

=>\(\frac{1}{2^2}\)x (\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{n^2}\))

Đặt A=\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{n^2}\)

Ta có:\(\frac{1}{2^2}\)<\(\frac{1}{1\cdot2}\)

         \(\frac{1}{3^2}\)<\(\frac{1}{2\cdot3}\)

.........\(\frac{1}{n^2}\)<\(\frac{1}{\left(n-1\right)\cdot n}\)

=>\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{n^2}\)<\(\frac{1}{1\cdot2}\)+\(\frac{1}{2\cdot3}\)+...+\(\frac{1}{\left(n-1\right)\cdot n}\)

=>A<1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{n-1}\)--\(\frac{1}{n}\)
=>\(\frac{1}{2^2}\)*A<\(\frac{1}{2^2}\)(1--\(\frac{1}{n}\))

=>\(\frac{1}{2^2}\)*A<\(\frac{1}{4}\)(1--\(\frac{1}{n}\))

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

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

Answer 4

Gọi tổng trên là S, ta có : 

S = 1/4^2+1/6^2+1/8^2+...+1/(2n)^2

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

S < 1/2^2.[1/1.2+1/2.3+...+1/(n-1).n]

S < 1/4.(1-1/2+1/2-1/3+...+1/(n-1)-1/n

S < 1/4.(1-1/n)

S < 1/4-1/4n <1/4

Suy ra , S <1/4