Sum of Squares of Binomial Coefficients Proof

Concept Overview

This problem tests the understanding of the Binomial Theorem and its applications in combinatorial identities. Specifically, it requires proving a fundamental identity involving the sum of squares of binomial coefficients. The proof often utilizes the concept of comparing coefficients of a polynomial derived from the Binomial Theorem, a technique closely related to Vandermonde's Identity.

Step 1: Recall the Binomial Theorem. The Binomial Theorem states that for any non-negative integer $n$ , the expansion of $(x+y)^n$ is given by:

(x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k

This formula allows us to express powers of binomials as sums involving binomial coefficients.

Step 2: Consider the expansion of $(1+x)^n$ . Using the Binomial Theorem with $y=1$ , we get the expansion of $(1+x)^n$ :

(1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{n}x^n

This expansion is crucial as it relates powers of $x$ to binomial coefficients.

Step 3: Consider the expansion of $(x+1)^n$ . Similarly, we can write the expansion of $(x+1)^n$ :

(x+1)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1} + \binom{n}{2}x^{n-2} + \dots + \binom{n}{n}

Note that $(x+1)^n = (1+x)^n$ , so these two expansions represent the same polynomial.

Step 4: Consider the product of the two expansions. Let's multiply the two expansions:

(1+x)^n \cdot (x+1)^n = \left( \sum_{i=0}^{n} \binom{n}{i} x^i \right) \left( \sum_{j=0}^{n} \binom{n}{j} x^{n-j} \right)

The product of these two polynomials will result in a new polynomial.

Step 5: Simplify the left side of the product. The left side of the equation can be simplified using the property $a^m \cdot b^m = (ab)^m$ :

(1+x)^n \cdot (x+1)^n = ((1+x)(x+1))^n = (1+x)^{2n}

This simplifies the problem to finding the coefficient of a specific term in the expansion of $(1+x)^{2n}$ .

Step 6: Expand $(1+x)^{2n}$ using the Binomial Theorem. The expansion of $(1+x)^{2n}$ is:

(1+x)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} x^k

We are interested in the coefficient of $x^n$ in this expansion, which is $\binom{2n}{n}$ .

Step 7: Determine the coefficient of $x^n$ in the product of the two series. When we multiply the two series from Step 2 and Step 3, the term with $x^n$ is obtained by summing products of terms where the powers of $x$ add up to $n$ .

\left( \binom{n}{0} + \binom{n}{1}x + \dots + \binom{n}{n}x^n \right) \left( \binom{n}{0}x^n + \binom{n}{1}x^{n-1} + \dots + \binom{n}{n} \right)

The coefficient of $x^n$ in this product is formed by terms $\binom{n}{i}x^i \cdot \binom{n}{j}x^{n-j}$ where $i+j=n$ . This means $j = n-i$ . So, the coefficient of $x^n$ is the sum of $\binom{n}{i} \binom{n}{n-i}$ for $i$ from $0$ to $n$ .

Step 8: Use the identity $\binom{n}{k} = \binom{n}{n-k}$ . We know that $\binom{n}{n-i} = \binom{n}{i}$ . Substituting this into the sum from Step 7:

\text{Coefficient of } x^n = \sum_{i=0}^{n} \binom{n}{i} \binom{n}{n-i} = \sum_{i=0}^{n} \binom{n}{i} \binom{n}{i} = \sum_{i=0}^{n} \binom{n}{i}^2

This sum represents $\binom{n}{0}^2 + \binom{n}{1}^2 + \dots + \binom{n}{n}^2$ .

Step 9: Equate the coefficients. By comparing the coefficient of $x^n$ from Step 6 (which is $\binom{2n}{n}$ ) and the coefficient of $x^n$ from Step 8 (which is $\sum_{i=0}^{n} \binom{n}{i}^2$ ), we arrive at the desired identity:

\sum_{i=0}^{n} \binom{n}{i}^2 = \binom{2n}{n}

Key Takeaways:

The Binomial Theorem is a powerful tool for expanding powers of binomials and understanding the structure of binomial coefficients.
Comparing coefficients of polynomials is a common technique for proving combinatorial identities.
The identity $\binom{n}{k} = \binom{n}{n-k}$ is frequently used in simplifying sums involving binomial coefficients.
The product $(1+x)^n \cdot (1+x)^n = (1+x)^{2n}$ provides a framework to relate sums of products of coefficients to a single binomial coefficient.

Answer: $\boxed{\sum_{i=0}^{n} \binom{n}{i}^2 = \binom{2n}{n}}$

Concept Overview

Key Takeaways:

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