Question

Let \( \alpha = \sum_{k=0}^{n} \left( \frac{\binom{n}{k}}{k+1} \right)^2 \) and \( \beta = \sum_{k=0}^{n-1} \left( \frac{\binom{n}{k} \binom{n}{k+1}}{k+2} \right) \).
If \( 5\alpha = 6\beta \), then \( n \) equals __________.

Answer 1

Correct Answer: 10

We are given:

\[ \alpha = \sum_{k=0}^{n} \frac{C_{k}^{n} \cdot C_{n-k}^{n}}{k + 1}. \]

This can be simplified using a known identity:

\[ \alpha = \frac{1}{n + 1} \sum_{k=0}^{n} C_{k}^{n+1} \cdot C_{n-k}^{n+1}. \]

Thus,

\[ \alpha = \frac{1}{n + 1} \cdot C_{2n+1}^{n+1}. \]

Next, we define \( \beta \) as follows:

\[ \beta = \sum_{k=0}^{n-1} \frac{C_{k}^{n} \cdot C_{n-k+1}^{n}}{k + 2}. \]

Using a similar identity, we can simplify this expression to:

\[ \beta = \frac{1}{n + 1} \cdot C_{2n+2}^{n+2}. \]

Now, to find the relationship between \( \beta \) and \( \alpha \), we compute \( \frac{\beta}{\alpha} \):

\[ \frac{\beta}{\alpha} = \frac{\frac{1}{n+1} \cdot C_{2n+2}^{n+2}}{\frac{1}{n+1} \cdot C_{2n+1}^{n+1}} = \frac{C_{2n+2}^{n+2}}{C_{2n+1}^{n+1}}. \]

This simplifies to:

\[ \frac{\beta}{\alpha} = \frac{n + 2}{n + 2} = \frac{5}{6}. \]

Thus, we find \( n = 10 \).