Solve the Following Questions

Question:

For $\mathrm{k} \in \mathrm{N}$, let

$\frac{1}{\alpha(\alpha+1)(\alpha+2) \ldots \ldots . .(\alpha+20)}=\sum_{\mathrm{K}=0}^{20} \frac{\mathrm{A}_{\mathrm{k}}}{\alpha+\mathrm{k}}$

where $\alpha>0$. Then the value of $100\left(\frac{\mathrm{A}_{14}+\mathrm{A}_{15}}{\mathrm{~A}_{13}}\right)^{2}$ is equal to

Solution:

$\frac{1}{\alpha(\alpha+1) \ldots . .(\alpha+20)}=\sum_{\mathrm{k}=0}^{20} \frac{\mathrm{A}_{\mathrm{k}}}{\alpha+\mathrm{k}}$

$\mathrm{A}_{14}=\frac{1}{(-14)(-13) \ldots . .(-1)(1) \ldots . .(6)}=\frac{1}{14 ! \cdot 6 !}$

$\mathrm{A}_{15}=\frac{1}{(-15)(-14) \ldots . .(-1)(1) \ldots . .(5)}=\frac{1}{15 ! \cdot 5 !}$

$\mathrm{A}_{13}=\frac{1}{(-13) \ldots . .(-1)(1) \ldots . .(7)}=\frac{-1}{13 ! \cdot 7 !}$

$\frac{\mathrm{A}_{14}}{\mathrm{~A}_{13}}=\frac{1}{14 ! \cdot 6 !} \times-13 ! \times 7 !=\frac{-7}{14}=-\frac{1}{2}$

$\frac{\mathrm{A}_{15}}{\mathrm{~A}_{13}}=-\frac{1}{15 ! \times 5 !} \times-13 ! \times 7 !=\frac{42}{15 \times 14}=\frac{1}{5}$

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