Question

[phạm phương thảo]

cho P = 1/2^2 + 1/4^2 + 1/6^2 + ... + 1/100^2, Chứng minh rằng P < 1/2

Answer 1

[Nguyễn Minh Quang]

ta có : 

\(P=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}=\frac{1}{4}\times\left(\frac{1}{1^2}+\frac{1}{2^2}+..+\frac{1}{50^2}\right)\)

\(< \frac{1}{4}\times\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)=\frac{1}{4}\times\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(=\frac{1}{4}\times\left(2-\frac{1}{50}\right)< \frac{1}{4}\times2=\frac{1}{2}\)

Vậy \(P< \frac{1}{2}\)