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

Given two expressions, \( \alpha \) and \( \beta \), which are sums of binomial coefficients, and a relation \( 5\alpha = 6\beta \), the objective is to determine the value of \( n \).

The expressions are defined as: \[ \alpha = \sum_{k=0}^{n} \frac{\binom{n}{k}^2}{k+1} \] \[ \beta = \sum_{k=0}^{n-1} \frac{\binom{n}{k} \binom{n}{k+1}}{k+2} \]

Concepts Utilized:

The simplification of \( \alpha \) and \( \beta \) will employ the following binomial coefficient identities:

1. Identity 1: \( \frac{1}{k+1} \binom{n}{k} = \frac{1}{n+1} \binom{n+1}{k+1} \)
2. Identity 2: \( \frac{1}{k+2} \binom{n}{k+1} = \frac{1}{n+1} \binom{n+1}{k+2} \)
3. Symmetry Identity: \( \binom{n}{k} = \binom{n}{n-k} \)
4. Vandermonde's Identity: \( \sum_{k=0}^{m} \binom{r}{k} \binom{s}{m-k} = \binom{r+s}{m} \). This is equivalent to finding the coefficient of \( x^m \) in \( (1+x)^r (1+x)^s \).

Following the simplification of \( \alpha \) and \( \beta \) into closed forms, the given equation will be solved for \( n \).

Solution Steps:

Step 1: Simplify \( \alpha \).

Starting with \( \alpha = \sum_{k=0}^{n} \frac{\binom{n}{k}^2}{k+1} = \sum_{k=0}^{n} \left(\frac{\binom{n}{k}}{k+1}\right) \binom{n}{k} \).

Applying Identity 1: \( \alpha = \sum_{k=0}^{n} \frac{1}{n+1} \binom{n+1}{k+1} \binom{n}{k} = \frac{1}{n+1} \sum_{k=0}^{n} \binom{n}{k} \binom{n+1}{k+1} \).

Let \( S_{\alpha} = \sum_{k=0}^{n} \binom{n}{k} \binom{n+1}{k+1} \). Using the symmetry identity, \( \binom{n+1}{k+1} = \binom{n+1}{n-k} \). Thus, \( S_{\alpha} = \sum_{k=0}^{n} \binom{n}{k} \binom{n+1}{n-k} \).

Applying Vandermonde's Identity with \( r=n, s=n+1, m=n \), we get \( S_{\alpha} = \binom{n + (n+1)}{n} = \binom{2n+1}{n} \).

Therefore, \( \alpha = \frac{1}{n+1} \binom{2n+1}{n} \).

Step 2: Simplify \( \beta \).

Starting with \( \beta = \sum_{k=0}^{n-1} \frac{\binom{n}{k} \binom{n}{k+1}}{k+2} = \sum_{k=0}^{n-1} \binom{n}{k} \left(\frac{\binom{n}{k+1}}{k+2}\right) \).

Applying Identity 2: \( \beta = \sum_{k=0}^{n-1} \binom{n}{k} \frac{1}{n+1} \binom{n+1}{k+2} = \frac{1}{n+1} \sum_{k=0}^{n-1} \binom{n}{k} \binom{n+1}{k+2} \).

Let \( S_{\beta} = \sum_{k=0}^{n-1} \binom{n}{k} \binom{n+1}{k+2} \). Using the symmetry identity, \( \binom{n+1}{k+2} = \binom{n+1}{n-k-1} \). Thus, \( S_{\beta} = \sum_{k=0}^{n-1} \binom{n}{k} \binom{n+1}{n-k-1} \).

Applying Vandermonde's Identity with \( r=n, s=n+1, m=n-1 \) (since \( n-k-1 = (n-1)-k \)), we get \( S_{\beta} = \binom{n + (n+1)}{n-1} = \binom{2n+1}{n-1} \).

Therefore, \( \beta = \frac{1}{n+1} \binom{2n+1}{n-1} \).

Step 3: Solve \( 5\alpha = 6\beta \) for \( n \).

Substituting the simplified forms: \( 5 \left( \frac{1}{n+1} \binom{2n+1}{n} \right) = 6 \left( \frac{1}{n+1} \binom{2n+1}{n-1} \right) \).

Canceling \( \frac{1}{n+1} \): \( 5 \binom{2n+1}{n} = 6 \binom{2n+1}{n-1} \).

Expanding using \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \): \( 5 \cdot \frac{(2n+1)!}{n!(n+1)!} = 6 \cdot \frac{(2n+1)!}{(n-1)!(n+2)!} \).

Canceling \( (2n+1)! \): \( \frac{5}{n!(n+1)!} = \frac{6}{(n-1)!(n+2)!} \).

Using \( n! = n \cdot (n-1)! \) and \( (n+2)! = (n+2) \cdot (n+1)! \): \( \frac{5}{n \cdot (n-1)! (n+1)!} = \frac{6}{(n-1)! (n+2)(n+1)!} \).

Canceling \( (n-1)! \) and \( (n+1)! \): \( \frac{5}{n} = \frac{6}{n+2} \).

Final Calculation and Result:

Cross-multiplying the equation \( \frac{5}{n} = \frac{6}{n+2} \):

\[ 5(n+2) = 6n \] \[ 5n + 10 = 6n \] \[ 10 = n \]

The value of \( n \) is determined to be 10.