Question

cho P=\(\dfrac{1}{5^2}+\dfrac{1}{7^2}+\dfrac{1}{9^2}+...+\dfrac{1}{201^2}+\dfrac{1}{203^2}\)  chứng tỏ rằng p<\(\dfrac{1}{6}\)

Answer 1

Lời giải:
$P< \frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{199.201}+\frac{1}{201.203}$

$P< \frac{1}{2}(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{199.201}+\frac{2}{201.203})$

$P< \frac{1}{2}(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{199}-\frac{1}{201}+\frac{1}{201}-\frac{1}{203})$
$P< \frac{1}{2}(\frac{1}{3}-\frac{1}{203})< \frac{1}{2}.\frac{1}{3}=\frac{1}{6}$