JEE Mains · Maths · STD 11 - 7. binomial theoram
Let \(\alpha=\sum_{k=0}^n\left(\frac{\left({ }^n C_k\right)^2}{k+1}\right)\) and \(\beta=\sum_{k=0}^{n-1}\left(\frac{{ }^n C_k{ }^n C_{k+1}}{k+2}\right)\). If \(5 \alpha=6 \beta\), then \(n\) equals
- A \(6\)
- B \(7\)
- C \(9\)
- D \(10\)
Answer & Solution
Correct Answer
(D) \(10\)
Step-by-step Solution
Detailed explanation
\(\sum_{k=0}^n \frac{{ }^n C_k \cdot{ }^n C_k}{k+1} \cdot \frac{n+1}{n+1} \) \( =\frac{1}{n+1} \sum_{k=0}^n{ }^{n+1} C_{k+1} \cdot{ }^n C_{n-k} \) \( =\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+1} \) \( =\sum_{k=0}^{n-1}{ }^n C_k \cdot \frac{{ }^n C_{k+1}}{k+2} \frac{n+1}{n+1} \)…
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