\[{\text{Prove that}}\] $$\sum\limits_{k = 0}^n {\frac{{\left( {2k} \right)!!}}{{\left( {2k + 1} \right)!!}}} = \frac{{2n + 3}}{2} \cdot {\rm B}\left( {\frac{1}{2},n + 2} \right) - 1.$$