Lời giải:
$A=\frac{1}{2^2}+\frac{1}{(2.2)^2}+\frac{1}{(2.3)^2}+...+\frac{1}{(2.50)^2}$
$=\frac{1}{2^2}(1+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2})$
$<\frac{1}{4}(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50})$
$=\frac{1}{4}(1+\frac{2-1}{1.2}+\frac{3-2}{2.3}+....+\frac{50-49}{49.50})$
$=\frac{1}{4}(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50})$
$=\frac{1}{4}(1+1-\frac{1}{50})< \frac{1}{4}(1+1)=\frac{1}{2}$ (đpcm)
A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\) + ... +\(\dfrac{1}{100^2}\)
A = \(\dfrac{1}{2^2}\)\((\)1+\(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) +...+\(\dfrac{1}{50^2}\)\()\) nhỏ hơn \(\dfrac{1}{4}\)\((\)\(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) +...+\(\dfrac{1}{49.50}\)\()\)
A = \(\dfrac{1}{4}\)\(\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
A=\(\dfrac{1}{4}\)\(\left(1+1-\dfrac{1}{50}\right)\)
A=\(\dfrac{99}{200}\) nhỏ hơn \(\dfrac{1}{2}\)
