To solve this problem, let's examine and interpret the left and right sides of the given identity:

Problem Statement Recap

We need to prove that: $\sum_{k=1}^n \binom{n}{k} \binom{n}{k-1} = \frac{\binom{2n+2}{n+1}}{2} - \binom{2n}{n}.$

Approach to the Solution

1. Understanding the Left-Hand Side (LHS): The left side involves a sum over the product of binomial coefficients, specifically: $\sum_{k=1}^n \binom{n}{k} \binom{n}{k-1}.$ To analyze this expression, we can try interpreting it combinatorially or using identities related to binomial coefficients.

2. Combinatorial Identity Approach: This sum has a known combinatorial interpretation, where it represents the ways to select subsets from two disjoint sets. A useful identity in this context is: $\sum_{k=0}^n \binom{n}{k} \binom{n}{k} = \binom{2n}{n}.$ However, since our expression involves $\binom{n}{k-1}$ rather than $\binom{n}{k}$, adjustments to this approach may be needed.

3. Using Generating Functions: Generating functions can sometimes simplify such sums, especially when dealing with binomial coefficients. The binomial theorem and related expansions may help.

4. Verifying the Right-Hand Side (RHS): The right side contains terms involving central binomial coefficients: $\frac{\binom{2n+2}{n+1}}{2} - \binom{2n}{n}.$ We may calculate each part separately and then compare it to our findings for the LHS.

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Questions for Further Exploration

1. What is the combinatorial meaning of $\binom{n}{k} \binom{n}{k-1}$ in terms of subsets?
2. Can we derive a general formula for sums involving products of binomial coefficients with different parameters?
3. How does the generating function method help in simplifying sums of binomial products?
4. How would this identity change if the range of $k$ extended beyond 1 to $n$?
5. Is there a recurrence relation that can be derived from this sum?

Tip

When working with binomial coefficient identities, it often helps to explore combinatorial interpretations, as these can reveal deeper insights into why certain sums and identities hold.