For $k \in N$, let $\frac{1}{\alpha(\alpha+1)(\alpha+2) \ldots(\alpha+20)}=\sum_{k=0}^{20} \frac{A_{

Question

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

Professor PC Thomas Classes · Answer

(9) : We have,\ $\frac{1}{\alpha(\alpha+1)(\alpha+2) \ldots .(\alpha+20)}$\ $=\sum_{k=0}^{20} \frac{A_{k}}{\alpha+k}, k \in N$\ $\Rightarrow \frac{1}{\alpha(\alpha+1)(\alpha+2) \ldots(\alpha+20)}$\ $=\frac{A_{0}}{\alpha}+\frac{A_{1}}{\alpha+1}+\frac{A_{2}}{\alpha+2}+\ldots+\frac{A_{20}}{\alpha+20}$\ $\Rightarrow \quad 1=A_{0}(\alpha+1)(\alpha+2) \ldots(\alpha+20)$\ $+A_{1}(\alpha)(\alpha+2) \ldots(\alpha+20)+\ldots+A_{20}(\alpha)$ $(\alpha+1) \ldots(\alpha+19)$\ Now, putting $\alpha=-13$ in above relation, we get $$ \begin{aligned} & A_{13}=\frac{1}{(-13)(-12)(-11) \ldots(-13+12)(-13+14)} \ & =-\frac{1}{\lfloor 13 \cdot \mid 7} \end{aligned} $$ Similarly, $$ A_{14}=\frac{1}{\begin{array}{l} (-14)(-13) \ldots(-14+13) \ (-14+15) \ldots(-14+20) \end{array}}=\frac{1}{14 \cdot 16} $$ and\ $A_{15}=\frac{1}{(-15)(-14) \ldots .(-15+20)}=-\frac{1}{\underline{15} \cdot \underline{15}}$\ Now, $\frac{A_{14}}{A_{13}}=\frac{-\lfloor 13floor 7}{\lfloor 14\lfloor 6}=-\frac{7}{14}=-\frac{1}{2}$;\ $\frac{A_{15}}{A_{13}}=\frac{\underline{13} \cdot \underline{7}}{\underline{15} \cdot \underline{5}}=\frac{1}{5}$\ Value of $100\left(\frac{A_{14}+A_{15}}{A_{13}}ight)^{2}=100\left(\frac{A_{14}}{A_{13}}+\frac{A_{15}}{A_{13}}ight)^{2}$\ $=100\left(-\frac{1}{2}+\frac{1}{5}ight)^{2}=100 	imes\left(\frac{-3}{10}ight)^{2}=9$.\