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 two expressions, \( \alpha \) and \( \beta \), involving sums of binomial coefficients, and a relation between them. We need to find the value of \( n \).

The expressions are: \[ \alpha = \sum_{k=0}^{n} \frac{\binom{n}{k}^2}{k+1} \] \[ \beta = \sum_{k=0}^{n-1} \frac{\binom{n}{k} \binom{n}{k+1}}{k+2} \] And the given relation is \( 5\alpha = 6\beta \).

Concept Used:

To simplify the expressions for \( \alpha \) and \( \beta \), we will use the following standard identities for binomial coefficients:

1. Absorption/Integration Identity 1: \( \frac{1}{k+1} \binom{n}{k} = \frac{1}{n+1} \binom{n+1}{k+1} \)
2. Absorption/Integration 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: \( \sum_{k=0}^{m} \binom{r}{k} \binom{s}{m-k} = \binom{r+s}{m} \). A useful application is to find the coefficient of \( x^m \) in the expansion of \( (1+x)^r (1+x)^s \).

After finding closed-form expressions for \( \alpha \) and \( \beta \), we will solve the given equation for \( n \).

Step-by-Step Solution:

Step 1: Simplify the expression for \( \alpha \).

We start with the definition of \( \alpha \):

\[ \alpha = \sum_{k=0}^{n} \frac{\binom{n}{k}^2}{k+1} = \sum_{k=0}^{n} \left(\frac{\binom{n}{k}}{k+1}\right) \binom{n}{k} \]

Using the identity \( \frac{1}{k+1} \binom{n}{k} = \frac{1}{n+1} \binom{n+1}{k+1} \):

\[ \alpha = \sum_{k=0}^{n} \frac{1}{n+1} \binom{n+1}{k+1} \binom{n}{k} = \frac{1}{n+1} \sum_{k=0}^{n} \binom{n}{k} \binom{n+1}{k+1} \]

To evaluate the sum, let's call it \( S_{\alpha} \). We use the symmetry identity \( \binom{n+1}{k+1} = \binom{n+1}{n+1-(k+1)} = \binom{n+1}{n-k} \).

\[ S_{\alpha} = \sum_{k=0}^{n} \binom{n}{k} \binom{n+1}{n-k} \]

This sum is in the form of Vandermonde's Identity, \( \sum_{k} \binom{r}{k} \binom{s}{m-k} = \binom{r+s}{m} \), with \( r=n, s=n+1, m=n \).

\[ S_{\alpha} = \binom{n + (n+1)}{n} = \binom{2n+1}{n} \]

Therefore, the expression for \( \alpha \) is:

\[ \alpha = \frac{1}{n+1} \binom{2n+1}{n} \]

Step 2: Simplify the expression for \( \beta \).

We start with the definition of \( \beta \):

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

Using the identity \( \frac{1}{k+2} \binom{n}{k+1} = \frac{1}{n+1} \binom{n+1}{k+2} \):

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

To evaluate this sum, let's call it \( S_{\beta} \). We use the symmetry identity \( \binom{n+1}{k+2} = \binom{n+1}{n+1-(k+2)} = \binom{n+1}{n-k-1} \).

\[ S_{\beta} = \sum_{k=0}^{n-1} \binom{n}{k} \binom{n+1}{n-k-1} \]

This sum is also in the form of Vandermonde's Identity, with \( r=n, s=n+1, m=n-1 \). Note that \( n-k-1 = (n-1)-k \).

\[ S_{\beta} = \binom{n + (n+1)}{n-1} = \binom{2n+1}{n-1} \]

Therefore, the expression for \( \beta \) is:

\[ \beta = \frac{1}{n+1} \binom{2n+1}{n-1} \]

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

Substitute the simplified expressions for \( \alpha \) and \( \beta \) into the equation:

\[ 5 \left( \frac{1}{n+1} \binom{2n+1}{n} \right) = 6 \left( \frac{1}{n+1} \binom{2n+1}{n-1} \right) \]

We can cancel the common factor \( \frac{1}{n+1} \) from both sides:

\[ 5 \binom{2n+1}{n} = 6 \binom{2n+1}{n-1} \]

Now, we expand the binomial coefficients using the factorial definition \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \):

\[ 5 \cdot \frac{(2n+1)!}{n!(2n+1-n)!} = 6 \cdot \frac{(2n+1)!}{(n-1)!(2n+1-(n-1))!} \] \[ 5 \cdot \frac{(2n+1)!}{n!(n+1)!} = 6 \cdot \frac{(2n+1)!}{(n-1)!(n+2)!} \]

Cancel the common term \( (2n+1)! \) from both sides:

\[ \frac{5}{n!(n+1)!} = \frac{6}{(n-1)!(n+2)!} \]

To simplify, we write the larger factorials in terms of smaller ones: \( n! = n \cdot (n-1)! \) and \( (n+2)! = (n+2) \cdot (n+1)! \).

\[ \frac{5}{n \cdot (n-1)! (n+1)!} = \frac{6}{(n-1)! (n+2)(n+1)!} \]

Cancel the common factors \( (n-1)! \) and \( (n+1)! \):

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

Final Computation & Result:

We solve the simple linear equation for \( n \) by cross-multiplication:

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

Thus, the value of \( n \) is 10.