Question

[Hoàng Bảo]

Cho $A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\mathrm{...}+\frac{1}{{100}^2}$. Chứng minh $A<\frac{1}{2}$.

Answer 1

[Nguyễn Thị Thương Hoài]

A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\) + ... + \(\dfrac{1}{100^2}\)

A = \(\dfrac{1}{2^2}\)  + \(\dfrac{1}{2^2}\).(\(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + ... + \(\dfrac{1}{50^2}\))

A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^2}\).(\(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + \(\dfrac{1}{4.4}\) + ... + \(\dfrac{1}{100.100}\))

A < \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\).(\(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ... + \(\dfrac{1}{49.50}\))

A < \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) .(\(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{49}\) - \(\dfrac{1}{50}\)

A < \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) ( 1 - \(\dfrac{1}{50}\))

A < \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{50}\) 

A < \(\dfrac{1}{2}\) - \(\dfrac{1}{50}\) < \(\dfrac{1}{2}\) (đpcm)