For k $\in$ N, let $${1 \over {\alpha (\alpha + 1)(\alpha + 2).........(\alpha + 20)}} = \sum\limits_{K = 0}^{20} {{{{A_k}} \over {\alpha + k}}} $$, where $\alpha > 0$. Then the value of $100{\left( {{{{A_{14}} + {A_{15}}} \over {{A_{13}}}}} \right)^2}$ is equal to _____________.

$${1 \over {\alpha (\alpha + 1)(\alpha + 2).........(\alpha + 20)}} = \sum\limits_{K = 0}^{20} {{{{A_k}} \over {\alpha + k}}} $$<br><br>$${A_{14}} = {1 \over {( - 14)( - 13)......( - 1)(1).......(6)}} = {1 \over {14!.6!}}$$<br><br>$${A_{15}} = {1 \over {( - 15)( - 14)......( - 1)(1).......(5)}} = {1 \over {15!.5!}}$$<br><br>${A_{13}} = {1 \over {( - 13)......( - 1)(1).......(7)}} = {1 \over {13!.7!}}$<br><br>$${{{A_{14}}} \over {{A_{13}}}} = {1 \over {14!.6!}} \times - 13! \times 7! = {{ - 7} \over {14}} = - {1 \over 2}$$<br><br>$${{{A_{15}}} \over {{A_{13}}}} = {1 \over {15! \times 5!}} \times - 13! \times 7! = {{42} \over {15 \times 14}} = {1 \over 5}$$<br><br>$$100{\left( {{{{A_{14}}} \over {{A_{13}}}} + {{{A_{15}}} \over {{A_{13}}}}} \right)^2} = 100{\left( { - {1 \over 2} + {1 \over 5}} \right)^2} = 9$$