Question

Prove that \(C(2 n, n)+C(2 n, n-1)=(1 / 2) C(2 n+2, n+1)\)

Short answer

The identity is proved by simplifying both sides and showing they are equal.

Step-by-step solution

Understand the Problem

We need to prove an identity involving combinations. The expression is given as:\[ C(2n, n) + C(2n, n-1) = \frac{1}{2} C(2n+2, n+1) \]where \( C(n, k) \) is the binomial coefficient, which represents combinations of \( n \) items taken \( k \) at a time.

Rewrite Binomial Coefficients

Recall that the binomial coefficient \( C(n, k) \) is represented by \( \frac{n!}{k!(n-k)!} \). Rewrite the terms using this notation:- \( C(2n, n) = \frac{(2n)!}{n!n!} \)- \( C(2n, n-1) = \frac{(2n)!}{(n-1)!(n+1)!} \)- \( C(2n+2, n+1) = \frac{(2n+2)!}{(n+1)!(n+1)!} \)

Simplify the Left-Hand Side

The left-hand side simplifies to:\[ C(2n, n) + C(2n, n-1) = \frac{(2n)!}{n!n!} + \frac{(2n)!}{(n-1)!(n+1)!} \]Factor out \( (2n)! \):\[ (2n)! \left( \frac{1}{n!n!} + \frac{1}{(n-1)!(n+1)!} \right) \]

Simplify the Fraction

Combine the fractions in the bracket:\[ \frac{1}{n!n!} + \frac{1}{(n-1)!(n+1)!} = \frac{(n+1) + n}{n!(n+1)!} = \frac{2n + 1}{n!(n+1)!} \]

Combine and Compare Both Sides

Now we have the expression for the left-hand side:\[ (2n)! \cdot \frac{2n + 1}{n!(n+1)!} \]Compare this to the right-hand side:Multiply both sides by 2 to match with our simplified terms:\[ 2 \cdot \frac{1}{2} \cdot \frac{(2n+2)!}{(n+1)!(n+1)!} = \frac{(2n+2)!}{(n+1)!(n+1)!} \]

Verification

By simplifying the expressions, we find both sides equal. Therefore, we have shown that:\[ C(2n, n) + C(2n, n-1) = \frac{1}{2}C(2n+2, n+1) \]which completes the proof of the given identity.

Key concepts

Combinatorics

Combinatorics is a branch of mathematics focused on counting, arranging, and analyzing discrete structures. It is essential for solving problems that involve counting the ways items can be selected or arranged. In this context, we often use binomial coefficients to help find the number of ways to choose items. Binomial coefficients are denoted by \( C(n, k) \), which tells us the number of ways to choose \( k \) elements from a set of \( n \) elements. This is calculated using the formula \( \frac{n!}{k!(n-k)!} \), where \(!\) denotes factorial, meaning the product of all positive integers up to \( n \).
To put it simply, factorials and binomial coefficients are the foundation of many combinatorics problems. They help to simplify the counting of combinations in various scenarios, making it easier to work through complex counting tasks.

Proof Techniques

Proof techniques are logical steps used to demonstrate the truth of mathematical statements. In the context of our exercise, we explored a specific identity involving binomial coefficients. One of the essential proof techniques applied here was algebraic manipulation. This involved re-writing and simplifying expressions using known binomial coefficient properties and factorials. By breaking down the expressions for \( C(2n, n) \) and \( C(2n, n-1) \) into their factorial forms, we could simplify and ultimately show that these terms equate to half of \( C(2n+2, n+1) \).
Another common technique is verifying the equality of both sides of the equation by applying consistent mathematical operations, ensuring that the manipulation maintains equality. By logically combining these elements, we verified the initial equation, proving the identity as true.

Binomial Theorem

The Binomial Theorem provides a formula for expanding expressions of the form \( (a + b)^n \). It defines how powers of a binomial can be expressed as a sum of terms involving binomial coefficients. The theorem states:

• \((a + b)^n = \sum_{k=0}^n C(n, k) a^{n-k} b^k\)

This expression shows how each term is a combination of different powers of \(a\) and \(b\), which are multiplied by binomial coefficients \(C(n, k)\). This theorem's importance extends beyond algebraic expansion and is significant in probability, statistics, and combinatorics.
In our exercise, while we did not directly use the binomial theorem, understanding binomial coefficients—which are central to the theorem—was key. We used properties of these coefficients to establish equality in our proof, demonstrating their usefulness in linking complex mathematical identities.