\(\sum\limits_{n = 1}^\infty {\frac{{{1^2}{{.5}^2}{{.9}^2}....{{\left( {4n - 3} \right)}^2}}}{{{4^2}{{.8}^2}{{.12}^2}....{{\left( {4n} \right)}^2}}}} \) ______ है।