Question

 Cho biểu thức:

A =1/2^2+1/4^2+1/6^2+.....+1/96^2+1/98^2.

Chứng tỏ A<1/2

Answer 1

\(A=\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{96^2}+\frac{1}{98^2}\)

\(A< \frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{95.97}+\frac{1}{97.99}\)

\(A< \frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}+\frac{1}{99}\)

\(A< 1-\frac{1}{99}\)

\(A< \frac{98}{99}\)