For $$k \in N$$, let $$\frac{1}{\alpha(\alpha+1)(\alpha+2)\ldots(\alpha+20)} = \sum_{K=0}^{20} \frac{A_K}{\alpha+k}$$, where $$\alpha > 0$$. Then the value of $$100\left(\frac{A_{14}+A_{15}}{A_{13}}\right)^2$$ is equal to ___.

Write the decomposition in the usual form

$$\frac{1}{\alpha(\alpha+1)\ldots(\alpha+20)}=\sum_{k=0}^{20}\frac{A_k}{\alpha+k},\qquad\alpha\gt 0.$$

To obtain a particular coefficient $$A_k$$, multiply both sides by $$(\alpha+k)$$ and put $$\alpha=-k$$:

$$A_k=\left.\frac{1}{\prod_{j=0,\,j\neq k}^{20}(\alpha+j)}\right|_{\alpha=-k} =\frac{1}{\prod_{j=0,\,j\neq k}^{20}(j-k)}.$$

Break the product into two parts, $$j\lt k$$ and $$j\gt k$$:

$$\prod_{j=0,\,j\neq k}^{20}(j-k)=\Bigl[\prod_{m=1}^{k}( -m)\Bigr] \Bigl[\prod_{m=1}^{20-k}m\Bigr] =(-1)^k k!\,(20-k)!.$$

Hence

$$A_k=\frac{1}{(-1)^k k!(20-k)!}=(-1)^k\frac{1}{k!(20-k)!}.$$

Compute the three required coefficients.

$$A_{13}=(-1)^{13}\frac{1}{13!\,7!}= -\frac{1}{13!\,7!},$$
$$A_{14}=(-1)^{14}\frac{1}{14!\,6!}= \frac{1}{14!\,6!},$$
$$A_{15}=(-1)^{15}\frac{1}{15!\,5!}= -\frac{1}{15!\,5!}.$$

Form the ratio

$$R=\frac{A_{14}+A_{15}}{A_{13}} =\frac{\dfrac{1}{14!\,6!}-\dfrac{1}{15!\,5!}} {-\dfrac{1}{13!\,7!}} =-\left(\frac{13!\,7!}{14!\,6!}-\frac{13!\,7!}{15!\,5!}\right).$$

Simplify each factor:

$$\frac{13!\,7!}{14!\,6!}=\frac{13!}{14!}\cdot\frac{7!}{6!} =\frac{1}{14}\cdot7=\frac12,$$
$$\frac{13!\,7!}{15!\,5!}=\frac{13!}{15!}\cdot\frac{7!}{5!} =\frac{1}{14\cdot15}\cdot42=\frac{42}{210}=\frac15.$$

Therefore

$$R=-\left(\frac12-\frac15\right)=-\left(\frac{5-2}{10}\right)=-\frac{3}{10}.$$

Square the ratio and multiply by $$100$$:

$$100\left(\frac{A_{14}+A_{15}}{A_{13}}\right)^2 =100\left(-\frac{3}{10}\right)^2 =100\cdot\frac{9}{100}=9.$$

Hence the required value is $$\mathbf{9}$$.