Question

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

Answer 1

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