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

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Binomial Theorem

For

k \in N

, let

\frac{1}{α ( α + 1 ) ( α + 2 ) \dots ( α + 20 )} = \sum_{k = 0}^{20} \frac{A _{k}}{α + k}

, where

α > 0

. Then the value of

100 (\frac{A _{14} + A _{15}}{A _{13}})^{2}

is equal to

Answer: 9

Multiplying both sides by

(α + k)

and letting

α \to - k

isolates

A_{k}

A_{k} = j = 0 j \neq = k \prod 20 \frac{1}{j - k} = \frac{1}{[ ( - 1 ) ^{k} k ! ] ( 20 - k )!} = \frac{( - 1 ) ^{k}}{k ! ( 20 - k )!}

A_{13} = - \frac{1}{13 ! 7 !}

A_{14} = \frac{1}{14 ! 6 !}

and

A_{15} = - \frac{1}{15 ! 5 !}

. Expressing the last two in terms of

\frac{1}{13 ! 7 !}

\frac{1}{14 ! 6 !} = \frac{7}{14} \cdot \frac{1}{13 ! 7 !} = \frac{1}{2} \cdot \frac{1}{13 ! 7 !}, \frac{1}{15 ! 5 !} = \frac{7 \times 6}{15 \times 14} \cdot \frac{1}{13 ! 7 !} = \frac{1}{5} \cdot \frac{1}{13 ! 7 !}

Hence

\frac{A _{14} + A _{15}}{A _{13}} = \frac{\frac{1}{2} - \frac{1}{5}}{- 1} = - \frac{3}{10}, 100 (\frac{A _{14} + A _{15}}{A _{13}})^{2} = 100 \times \frac{9}{100} = 9

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Integer answer