Tổng quát: $1-\frac{1}{x^2}=\frac{x^2-1}{x^2}=\frac{(x-1)(x+1)}{x^2}$
$⇒B=(1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2})...(1-\frac{1}{2028^2})$
$=\frac{1.3}{2^2}.\frac{2.4}{3^2}.....\frac{2027.2029}{2028^2}$
$=\frac{1}2.\frac{2029}{2028}$
$=\frac{2029}{4056}$
Vậy $B=\frac{2029}{4056}$