Question

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 _______.

Answer 1

Answer: 9

$\frac{1}{\alpha (\alpha +1)(\alpha +2)\dots \dots (\alpha +20)}=\sum _{K=0}^{20}\frac{A_k}{\alpha +k}$
$A_{14}=\frac{1}{(-14)(-13)......(-1)(1)\dots (6)}=\frac{1}{14!\cdot 6!}$
$A_{15}=\frac{1}{(-15)(-14)\dots \dots (-1)(1)\dots (5)}=\frac{1}{15!\cdot 5!}$
$A_{13}=\frac{1}{(-13)\dots \dots (-1)(1)\dots \dots (7)}=\frac{-1}{13!\cdot 7!}$
$\frac{A_{14}}{A_{13}}=\frac{1}{14!\cdot 6!}\times -13!\times 7!=\frac{-7}{14}=-\frac{1}{2}$
$\frac{A_{15}}{A_{13}}=-\frac{1}{15!\times 5!}\times -13!\times 7!=\frac{42}{15\times 14}=\frac{1}{5}$
$100{\left(\frac{A_{14}}{A_{13}}+\frac{A_{15}}{A_{13}}\right)}^2=100{\left(-\frac{1}{2}+\frac{1}{5}\right)}^2=9$