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

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 $n = 10$.