JEE Mains · Maths · STD 11 - 6. permutation and combination
For \(k \in N\), let \(\frac{1}{\alpha(\alpha+1)(\alpha+2) \ldots(\alpha+20)}=\sum_{k=0}^{20} \frac{A_{k}}{a+k}\), where \(a\,>\,0\). Then the value of \(100\left(\frac{A_{14}+A_{15}}{A_{13}}\right)^{2}\) is equal to \(....\)
- A \(9\)
- B \(27\)
- C \(3\)
- D \(81\)
Answer & Solution
Correct Answer
(A) \(9\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{\alpha(\alpha+1) \ldots(\alpha+20)}=\sum_{k=0}^{20} \frac{A_{k}}{\alpha+k}\) \(A_{14}=\frac{1}{(-14)(-13) \ldots(-1)(1) \ldots(6)}=\frac{1}{14 ! \cdot 6 !}\) \(A_{15}=\frac{1}{(-15)(-14) \ldots(-1)(1) \ldots(5)}=\frac{-1}{15 ! 5 !}\)…
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