Key Concepts and Formulae

This problem involves simplifying sums of binomial coefficients. The core identities we will utilize are:

1. Identity for Binomial Coefficients with Denominator: $\frac{{ }^n C_r}{r+1} = \frac{1}{n+1} { }^{n+1} C_{r+1}$

   • Explanation: This identity allows us to combine the binomial coefficient with the denominator, transforming a term like $\frac{{ }^n C_r}{r+1}$ into a single binomial coefficient of a higher order. This is derived from the definition of binomial coefficients: $\frac{{ }^n C_r}{r+1} = \frac{n!}{r!(n-r)! (r+1)} = \frac{n!}{(r+1)!(n-r)!}$ And $\frac{1}{n+1} { }^{n+1} C_{r+1} = \frac{1}{n+1} \frac{(n+1)!}{(r+1)!(n+1-(r+1))!} = \frac{1}{n+1} \frac{(n+1)n!}{(r+1)!(n-r)!} = \frac{n!}{(r+1)!(n-r)!}$ Hence, the identity holds.

2. Symmetry Property of Binomial Coefficients: ${ }^n C_k = { }^n C_{n-k}$

   • Explanation: This property states that choosing $k$ items from $n$ is the same as choosing $n-k$ items to leave behind. It's often used to manipulate indices to fit standard summation forms.

3. Vandermonde's Identity (Coefficient Extraction from Polynomial Product): $\sum_{k=0}^r { }^m C_k { }^n C_{r-k} = { }^{m+n} C_r$

   • Explanation: This identity arises from considering the coefficient of $x^r$ in the expansion of $(1+x)^m (1+x)^n = (1+x)^{m+n}$. When multiplying two polynomials, the coefficient of $x^r$ is obtained by summing products of coefficients where the powers of $x$ add up to $r$. For example, the $x^k$ term from $(1+x)^m$ combines with the $x^{r-k}$ term from $(1+x)^n$.

4. Ratio of Consecutive Binomial Coefficients: $\frac{{ }^N C_R}{{ }^N C_{R-1}} = \frac{N-R+1}{R}$

   • Explanation: This is a useful formula for simplifying ratios of binomial coefficients that share the same upper index ($N$) and whose lower indices ($R$ and $R-1$) are consecutive.

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Step-by-Step Solution

1. Evaluation of $\alpha$

We are given \alpha=\sum_\limits{k=0}^n\left(\frac{\left({ }^n C_k\right)^2}{k+1}\right).

• Step 1.1: Apply the denominator identity. $\alpha = \sum_{k=0}^n \left( { }^n C_k \cdot \frac{{ }^n C_k}{k+1} \right)$ Using the identity $\frac{{ }^n C_k}{k+1} = \frac{1}{n+1} { }^{n+1} C_{k+1}$, we substitute this into the expression for $\alpha$: $\alpha = \sum_{k=0}^n \left( { }^n C_k \cdot \frac{1}{n+1} { }^{n+1} C_{k+1} \right)$ $\alpha = \frac{1}{n+1} \sum_{k=0}^n { }^n C_k { }^{n+1} C_{k+1}$

  • Why this step? The denominator $(k+1)$ makes direct application of Vandermonde's Identity difficult. This transformation eliminates the denominator by promoting one of the binomial coefficients to a higher order, setting the stage for Vandermonde's Identity.

• Step 1.2: Apply the symmetry property. Now, we use the symmetry property ${ }^n C_k = { }^n C_{n-k}$ for the first binomial coefficient: $\alpha = \frac{1}{n+1} \sum_{k=0}^n { }^n C_{n-k} { }^{n+1} C_{k+1}$

  • Why this step? This manipulation helps to match the form required for Vandermonde's Identity, where the sum of the lower indices of the two binomial coefficients is constant.

• Step 1.3: Apply Vandermonde's Identity. The sum $\sum_{k=0}^n { }^n C_{n-k} { }^{n+1} C_{k+1}$ is equivalent to finding the coefficient of $x^{n+1}$ in the product of $(1+x)^n$ and $(1+x)^{n+1}$. Let $i = n-k$. As $k$ goes from $0$ to $n$, $i$ goes from $n$ to $0$. The sum becomes: $\sum_{i=0}^n { }^n C_i { }^{n+1} C_{(n-i)+1} = \sum_{i=0}^n { }^n C_i { }^{n+1} C_{n-i+1}$ Here, $m=n$, $p=n+1$, and the constant sum of the lower indices is $i + (n-i+1) = n+1$. According to Vandermonde's Identity, this sum equals ${ }^{n+(n+1)} C_{n+1}$. Therefore, $\sum_{k=0}^n { }^n C_{n-k} { }^{n+1} C_{k+1} = { }^{2n+1} C_{n+1}$ Substituting this back into the expression for $\alpha$: $\alpha = \frac{1}{n+1} { }^{2n+1} C_{n+1}$

  • Why this step? Vandermonde's Identity is the key to simplifying this type of sum of products of binomial coefficients into a single, compact binomial coefficient.

2. Evaluation of $\beta$}

We are given \beta=\sum_\limits{k=0}^{n-1}\left(\frac{{ }^n C_k{ }^n C_{k+1}}{k+2}\right).

• Step 2.1: Apply the denominator identity. $\beta = \sum_{k=0}^{n-1} \left( { }^n C_k \cdot \frac{{ }^n C_{k+1}}{k+2} \right)$ Using the identity $\frac{{ }^n C_r}{r+1} = \frac{1}{n+1} { }^{n+1} C_{r+1}$, we set $r=k+1$. So $r+1 = k+2$. Thus, $\frac{{ }^n C_{k+1}}{k+2} = \frac{1}{n+1} { }^{n+1} C_{(k+1)+1} = \frac{1}{n+1} { }^{n+1} C_{k+2}$. Substitute this into the expression for $\beta$: $\beta = \sum_{k=0}^{n-1} \left( { }^n C_k \cdot \frac{1}{n+1} { }^{n+1} C_{k+2} \right)$ $\beta = \frac{1}{n+1} \sum_{k=0}^{n-1} { }^n C_k { }^{n+1} C_{k+2}$

  • Why this step? Similar to $\alpha$, this transforms the term with a denominator into a single binomial coefficient, preparing it for Vandermonde's Identity.

• Step 2.2: Apply the symmetry property. Now, we use the symmetry property ${ }^n C_k = { }^n C_{n-k}$ for the first binomial coefficient: $\beta = \frac{1}{n+1} \sum_{k=0}^{n-1} { }^n C_{n-k} { }^{n+1} C_{k+2}$

  • Why this step? Again, this helps align the terms for a clear application of Vandermonde's Identity.

• Step 2.3: Apply Vandermonde's Identity. The sum $\sum_{k=0}^{n-1} { }^n C_{n-k} { }^{n+1} C_{k+2}$ is equivalent to finding the coefficient of $x^{n+2}$ in the product of $(1+x)^n$ and $(1+x)^{n+1}$. Let $i = n-k$. As $k$ goes from $0$ to $n-1$, $i$ goes from $n$ to $1$. The sum becomes: $\sum_{i=1}^n { }^n C_i { }^{n+1} C_{(n-i)+2} = \sum_{i=1}^n { }^n C_i { }^{n+1} C_{n-i+2}$ Here, $m=n$, $p=n+1$, and the constant sum of the lower indices is $i + (n-i+2) = n+2$. The Vandermonde's Identity requires the sum to typically go from $i=0$ to $n$. Let's check the $i=0$ term: ${ }^n C_0 { }^{n+1} C_{n+2} = 1 \cdot 0 = 0$ (since $n+2 > n+1$ for $n \ge 1$, which is implied as the upper limit of summation for $\alpha$ is $n$). Therefore, including the $i=0$ term does not change the sum. So, we can write the sum as $\sum_{i=0}^n { }^n C_i { }^{n+1} C_{n-i+2}$, which by Vandermonde's Identity equals ${ }^{n+(n+1)} C_{n+2}$. Therefore, $\sum_{k=0}^{n-1} { }^n C_{n-k} { }^{n+1} C_{k+2} = { }^{2n+1} C_{n+2}$ Substituting this back into the expression for $\beta$: $\beta = \frac{1}{n+1} { }^{2n+1} C_{n+2}$

  • Why this step? This is again the application of Vandermonde's Identity to simplify the sum. The careful consideration of summation limits ensures accuracy.

3. Finding the Ratio $\beta/\alpha$ and Solving for $n$

We are given the condition $5 \alpha = 6 \beta$, which implies $\frac{\beta}{\alpha} = \frac{5}{6}$.

• Step 3.1: Calculate the ratio $\beta/\alpha$. $\frac{\beta}{\alpha} = \frac{\frac{1}{n+1} { }^{2n+1} C_{n+2}}{\frac{1}{n+1} { }^{2n+1} C_{n+1}}$ $\frac{\beta}{\alpha} = \frac{{ }^{2n+1} C_{n+2}}{{ }^{2n+1} C_{n+1}}$ Using the identity for the ratio of consecutive binomial coefficients $\frac{{ }^N C_R}{{ }^N C_{R-1}} = \frac{N-R+1}{R}$: Here, $N = 2n+1$ and $R = n+2$. The term ${ }^N C_{R-1}$ corresponds to ${ }^{2n+1} C_{n+1}$. $\frac{{ }^{2n+1} C_{n+2}}{{ }^{2n+1} C_{n+1}} = \frac{(2n+1) - (n+2) + 1}{n+2}$ $\frac{\beta}{\alpha} = \frac{2n+1-n-2+1}{n+2} = \frac{n}{n+2}$

  • Why this step? This identity efficiently simplifies the ratio of two binomial coefficients, which would otherwise involve more complex factorial manipulations.

• Step 3.2: Solve for $n$. We have $\frac{\beta}{\alpha} = \frac{n}{n+2}$. Equating this to the given condition: $\frac{n}{n+2} = \frac{5}{6}$ Cross-multiply: $6n = 5(n+2)$ $6n = 5n + 10$ $6n - 5n = 10$ $n = 10$

  • Why this step? This is the final algebraic step to determine the value of $n$ based on the derived relationship between $\alpha$ and $\beta$.

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Tips and Common Mistakes

• Denominator Identity: Always be on the lookout for terms like $\frac{{ }^n C_k}{k+1}$ or $\frac{{ }^n C_k}{k+2}$. These are strong indicators that the identity $\frac{{ }^n C_r}{r+1} = \frac{1}{n+1} { }^{n+1} C_{r+1}$ should be used. Forgetting the $\frac{1}{n+1}$ factor is a common error.
• Vandermonde's Identity: Ensure correct mapping of indices. The sum of the lower indices of the two binomial coefficients must be constant ($k + (r-k) = r$). Also, remember to handle the limits of summation carefully, especially when extending them (e.g., adding zero terms).
• Algebraic Errors: Simple arithmetic or algebraic mistakes when solving the final equation for $n$ can lead to an incorrect answer. Double-check your calculations.
• Understanding the "Why": Merely memorizing formulas isn't enough. Understanding why each identity works (e.g., Vandermonde's as a coefficient of a product) helps in applying them correctly and tackling variations of problems.

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Summary

This problem beautifully demonstrates the power of binomial coefficient identities in simplifying complex sums. By strategically applying the identity $\frac{{ }^n C_r}{r+1} = \frac{1}{n+1} { }^{n+1} C_{r+1}$ to eliminate the denominators, followed by the symmetry property and Vandermonde's Identity, we transformed the given sums into single binomial coefficients. The final step involved using the ratio identity for binomial coefficients and solving a simple linear equation to find $n$. The key takeaway is the importance of recognizing and applying these fundamental combinatorial identities to simplify seemingly intractable expressions.