For 0, let 1(+1)(+2)(+20) = _k=0^20(A_k)/(+k).

Question

For $\alpha > 0$, let $\dfrac{1}{\alpha(\alpha+1)(\alpha+2)\cdots(\alpha+20)} = \displaystyle\sum_{k=0}^{20}\frac{A_k}{\alpha+k}$. Then a certain combination of $A_{13}, A_{14}, A_{15}$ equals:

Answer: 9

Accepted Answer

By the Heaviside cover-up rule, $A_k = \dfrac{(-1)^k}{k!(20-k)!}$, giving
$$A_{13} = \frac{-1}{13!\,7!}, \qquad A_{14}=\frac{1}{14!\,6!}, \qquad A_{15}=\frac{-1}{15!\,5!}.$$

$$\frac{A_{14}}{A_{13}} = -\frac{13!7!}{14!6!} = -\frac{7}{14}=-\frac12, \qquad \frac{A_{15}}{A_{13}} = \frac{13!7!}{15!5!} = \frac{42}{15\times14}=\frac15.$$

The question asks for $100\left(\dfrac{A_{14}}{A_{13}}+\dfrac{A_{15}}{A_{13}}\right)^2$:

$$100\left(-\frac12+\frac15\right)^2 = 100\left(-\frac{3}{10}\right)^2 = 9.$$

Solutions · 1