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

We have,

$\sum _{k=0}^{20}\left(\frac{A_k}{\alpha +k}\right)=\frac{1}{\alpha \left(\alpha +1\right)\dots \dots \left(\alpha +20\right)}$

$\Rightarrow \frac{A_0}{\alpha }+\frac{A_1}{\alpha +1}+...+\frac{A_{20}}{\alpha +20}=\frac{1}{\alpha \left(\alpha +1\right)\dots \dots \left(\alpha +20\right)}$

$\Rightarrow \frac{A_0\left(\alpha +1\right)\left(\alpha +2\right)\dots \dots \left(\alpha +19\right)\left(\alpha +20\right)+A_2\alpha \left(\alpha +2\right)\dots \dots \left(\alpha +19\right)\left(\alpha +20\right)+...+A_{20}\alpha \left(\alpha +1\right)\left(\alpha +2\right)\dots \dots \left(\alpha +19\right)}{\alpha \left(\alpha +1\right)\left(\alpha +2\right)\dots \dots \left(\alpha +20\right)}=\frac{1}{\alpha \left(\alpha +1\right)\left(\alpha +2\right)\dots \dots \left(\alpha +20\right)}$

$\Rightarrow A_0\left[\left(\alpha +1\right)\left(\alpha +2\right)\dots \dots \left(\alpha +19\right)\left(\alpha +20\right)\right]+A_2\left[\alpha \left(\alpha +2\right)\dots \dots \left(\alpha +19\right)\left(\alpha +20\right)\right]+...+A_{20}\left[\alpha \left(\alpha +1\right)\left(\alpha +2\right)\dots \dots \left(\alpha +19\right)\right]=1 ....\left(i\right)$

Hence, coefficient of $A_{13}$ is

$\alpha \left(\alpha +1\right)\dots \left(\alpha +12\right)\left(\alpha +14\right)\dots \left(\alpha +19\right)\left(\alpha +20\right)$.

Similarly, coefficient of $A_{14}$ is

$\alpha \left(\alpha +1\right)\dots \left(\alpha +13\right)\left(\alpha +15\right)\dots \left(\alpha +19\right)\left(\alpha +20\right)$

And, coefficient of $A_{15}$ is

$\alpha \left(\alpha +1\right)\dots \left(\alpha +14\right)\left(\alpha +16\right)\dots \left(\alpha +19\right)\left(\alpha +20\right)$

Then, to find $A_{14}$, put $\alpha =-14$ in $\left(i\right)$, then we get

$A_{14}=\frac{1}{(-14)(-13)\dots ..(-1)\left(1\right)\dots ..\left(6\right)}=\frac{1}{14!·6!}$

Similarly, putting $\alpha =-15$, we get

$A_{15}=\frac{1}{(-15)(-14)\dots \dots (-1)\left(1\right)\dots ..\left(5\right)}=-\frac{1}{15!·5!}$

And, putting $\alpha =-13$, we get

$A_{13}=\frac{1}{(-13)\dots \dots (-1)\left(1\right)\dots ..\left(7\right)}=\frac{-1}{13!·7!}$

Hence,

$\frac{A_{14}}{{ A}_{13}}=\frac{1}{14!·6!}\times \left(-13!\times 7!\right)=\frac{-7}{14}=-\frac{1}{2}$

$\frac{A_{15}}{{ A}_{13}}=\left(-\frac{1}{15!\times 5!}\right)\times \left(-13!\times 7!\right)=\frac{42}{15\times 14}=\frac{1}{5}$

Then,

$100{\left(\frac{A_{14}}{{ A}_{13}}+\frac{A_{15}}{{ A}_{13}}\right)}^2=100{\left(-\frac{1}{2}+\frac{1}{5}\right)}^2=9$