Đáp án+Giải thích các bước giải:
$A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{50^2}\\>\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{50.51}\\=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+..+\dfrac{1}{50}-\dfrac{1}{51}\\=\dfrac{1}{5}-\dfrac{1}{51}\\=\dfrac{46}{255}>\dfrac{45}{255}=\dfrac{3}{17}\\=>A>\dfrac{3}{17}$