For a 0,let frac{1}{a(a+1)(a+2) ldots(a+20) | vedclass

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(A) Using partial fraction decomposition,we have $A_k = \lim_{a 	o -k} \frac{a+k}{a(a+1)\ldots(a+20)}$.$A_k = \frac{1}{(-k)(-k+1)\ldots(-1)(1)(2)\ldo

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) Using partial fraction decomposition,we have $A_k = \lim_{a 	o -k} \frac{a+k}{a(a+1)\ldots(a+20)}$.$A_k = \frac{1}{(-k)(-k+1)\ldots(-1)(1)(2)\ldots(20-k)} = \frac{1}{(-1)^k k! (20-k)!}$.Thus,$A_k = \frac{(-1)^k}{k!(20-k)!}$.We need to calculate $\frac{A_{14}+A_{15}}{A_{13}}$.$A_{14} = \frac{(-1)^{14}}{14!6!} = \frac{1}{14!6!}$.$A_{15} = \frac{(-1)^{15}}{15!5!} = -\frac{1}{15!5!}$.$A_{13} = \frac{(-1)^{13}}{13!7!} = -\frac{1}{13!7!}$.$\frac{A_{14}}{A_{13}} = \frac{1}{14!6!} 	imes (-13!7!) = -\frac{7}{14} = -\frac{1}{2}$.$\frac{A_{15}}{A_{13}} = -\frac{1}{15!5!} 	imes (-13!7!) = \frac{7 	imes 6}{15 	imes 14} = \frac{42}{210} = \frac{1}{5}$.Therefore,$100\left(\frac{A_{14}}{A_{13}} + \frac{A_{15}}{A_{13}}ight)^2 = 100\left(-\frac{1}{2} + \frac{1}{5}ight)^2 = 100\left(-\frac{3}{10}ight)^2 = 100 	imes \frac{9}{100} = 9$.

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

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