JEE Main 2021 (Online) 20th July Evening ShiftQuestion 1
For $k\in\mathbf{N}$, let $\dfrac{1}{\alpha(\alpha+1)(\alpha+2)\cdots(\alpha+20)}=\displaystyle\sum_{k=0}^{20}\dfrac{A_k}{\alpha+k}$, where $\alpha>0$. Then the value of $100\left(\dfrac{A_{14}+A_{15}}{A_{13}}\right)^2$ is equal to $\_\_\_\_\_\_\_\_$ .
Sequence and Series · Miscellaneous
Answer$9$
Solution
Formulas used. (1) Cover-up rule: $A_k=\frac{1}{\prod_{j\ne k}(j-k)}$ (put $\alpha=-k$ in the rest of the denominator)Step 1 - Find $A_k$. For $j<k$ the factors $(j-k)$ are $-k,\dots,-1$; for $j>k$ they are $1,\dots,20-k$: $$A_k=\frac{1}{(-1)^k\,k!\,(20-k)!}=\frac{(-1)^k}{k!\,(20-k)!}.$$ Step 2 - Ratios. $$\frac{A_{14}}{A_{13}}=-\frac{13!\,7!}{14!\,6!}=-\frac{7}{14}=-\frac12,\qquad \frac{A_{15}}{A_{13}}=\frac{13!\,7!}{15!\,5!}=\frac{7\cdot6}{15\cdot14}=\frac15 .$$ Step 3. $$\frac{A_{14}+A_{15}}{A_{13}}=-\frac12+\frac15=-\frac3{10},\qquad 100\cdot\frac{9}{100}=9 .$$ Answer: 9