Key Concept: Sum of Squares of Binomial Coefficients

The problem requires us to evaluate a sum involving the squares of binomial coefficients. This type of sum is a direct application of a specific and important identity derived from the Binomial Theorem. The fundamental identity states: $\sum_{k=0}^{n} \left( \binom{n}{k} \right)^2 = \binom{2n}{n}$ This identity is a special case of Vandermonde's Identity and is crucial for simplifying such sums.

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Step-by-Step Working with Explanations

1. Analyze the Given Sum The sum we need to evaluate is: $S = \sum_{k = 0}^{20} {\left( {{}^{20}{C_k}} \right)^2}$ Let's expand a few terms to better understand its structure: $S = {\left( {{}^{20}{C_0}} \right)^2} + {\left( {{}^{20}{C_1}} \right)^2} + {\left( {{}^{20}{C_2}} \right)^2} + \dots + {\left( {{}^{20}{C_{20}}} \right)^2}$ Why this step is taken: Expanding the sum helps us visualize the pattern and confirm that it indeed involves the squares of binomial coefficients where the upper index is constant. It sets the stage for matching it with a known identity.

2. Identify 'n' for the Identity Comparing the given sum $\sum_{k = 0}^{20} {\left( {{}^{20}{C_k}} \right)^2}$ with the general form of the identity $\sum_{k=0}^{n} \left( \binom{n}{k} \right)^2$, we can clearly see that the value of $n$ in our problem is $20$. $n = 20$ Why this step is taken: This is a direct mapping from the problem's specific sum to the general form of the identity. Identifying the correct value of 'n' is the first critical step to correctly apply the formula.

3. Apply the Identity Now, substitute the identified value of $n=20$ into the identity $\sum_{k=0}^{n} \left( \binom{n}{k} \right)^2 = \binom{2n}{n}$: $\sum_{k=0}^{20} \left( {{}^{20}{C_k}} \right)^2 = {}^{2 \times 20}{C_{20}}$ $\sum_{k=0}^{20} \left( {{}^{20}{C_k}} \right)^2 = {}^{40}{C_{20}}$ Why this step is taken: This is the core application of the identity. Instead of performing a lengthy and complex calculation of each squared binomial coefficient, this identity allows us to directly compute the entire sum by forming a single, new binomial coefficient. This dramatically simplifies the problem.

4. Match with the Given Options The calculated value for the sum is ${}^{40}{C_{20}}$. Let's review the provided options: (A) ${}^{40}{C_{21}}$ (B) ${}^{40}{C_{19}}$ (C) ${}^{40}{C_{20}}$ (D) ${}^{41}{C_{20}}$ Our result perfectly matches option (C). Why this step is taken: This confirms that our application of the identity and subsequent calculation led to one of the correct choices provided in the problem.

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Derivation of the Key Identity (for Deeper Understanding)

The identity $\sum_{k=0}^{n} \left( \binom{n}{k} \right)^2 = \binom{2n}{n}$ can be elegantly derived by considering the coefficient of $x^n$ in the expansion of $(1+x)^n (x+1)^n$.

We know from the Binomial Theorem that: $(1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{k}x^k + \dots + \binom{n}{n}x^n \quad \text{(Equation 1)}$ And also: $(x+1)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1} + \binom{n}{2}x^{n-2} + \dots + \binom{n}{n-k}x^k + \dots + \binom{n}{n}x^0 \quad \text{(Equation 2)}$ A key property of binomial coefficients is symmetry: $\binom{n}{k} = \binom{n}{n-k}$. Using this property, we can rewrite Equation 2 as: $(x+1)^n = \binom{n}{n}x^n + \binom{n}{n-1}x^{n-1} + \dots + \binom{n}{k}x^{n-k} + \dots + \binom{n}{0}x^0$ Let's call the second expansion from Equation 2 (with $C(n,k)$ instead of $C(n,n-k)$): $(x+1)^n = \binom{n}{n}x^n + \binom{n}{n-1}x^{n-1} + \dots + \binom{n}{1}x^1 + \binom{n}{0}x^0$ Now, consider the product of these two binomial expansions: $(1+x)^n (x+1)^n = (1+x)^{2n}$ The coefficient of $x^n$ in the expansion of $(1+x)^{2n}$ is directly $\binom{2n}{n}$ by the Binomial Theorem.

Alternatively, to find the coefficient of $x^n$ in the product of Equation 1 and the rewritten Equation 2, we multiply terms such that their powers of $x$ sum to $n$: Coefficient of $x^n$ = $(\text{coeff of } x^0 \text{ in Eq 1}) \times (\text{coeff of } x^n \text{ in Eq 2}) +$ $\qquad \qquad \qquad (\text{coeff of } x^1 \text{ in Eq 1}) \times (\text{coeff of } x^{n-1} \text{ in Eq 2}) + \dots +$ $\qquad \qquad \qquad (\text{coeff of } x^n \text{ in Eq 1}) \times (\text{coeff of } x^0 \text{ in Eq 2})$

This gives us: $\binom{n}{0}\binom{n}{n} + \binom{n}{1}\binom{n}{n-1} + \binom{n}{2}\binom{n}{n-2} + \dots + \binom{n}{n}\binom{n}{0}$ Using the symmetry property $\binom{n}{k} = \binom{n}{n-k}$, we can rewrite each term: $\binom{n}{0}\binom{n}{0} + \binom{n}{1}\binom{n}{1} + \binom{n}{2}\binom{n}{2} + \dots + \binom{n}{n}\binom{n}{n}$ Which simplifies to: $\left( \binom{n}{0} \right)^2 + \left( \binom{n}{1} \right)^2 + \left( \binom{n}{2} \right)^2 + \dots + \left( \binom{n}{n} \right)^2 = \sum_{k=0}^{n} \left( \binom{n}{k} \right)^2$ By equating the two ways of finding the coefficient of $x^n$ in $(1+x)^{2n}$, we prove the identity: $\sum_{k=0}^{n} \left( \binom{n}{k} \right)^2 = \binom{2n}{n}$

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Tips and Common Mistakes

• Memorize Key Identities: For competitive exams, direct application of identities saves valuable time. This particular identity is a frequent guest in problems involving binomial coefficients.
• Vandermonde's Identity: Understand that the identity used here is a special case of the more general Vandermonde's Identity: $\sum_{k=0}^{r} \binom{m}{k}\binom{n}{r-k} = \binom{m+n}{r}$. If you set $m=n$ and $r=n$, it reduces to $\sum_{k=0}^{n} \binom{n}{k}\binom{n}{n-k} = \binom{2n}{n}$, which, using $\binom{n}{n-k}=\binom{n}{k}$, becomes $\sum_{k=0}^{n} \left( \binom{n}{k} \right)^2 = \binom{2n}{n}$.
• Symmetry Property: The property $\binom{n}{k} = \binom{n}{n-k}$ is fundamental and is used implicitly in many binomial coefficient problems. It means that the coefficient of $x^k$ in $(1+x)^n$ is the same as the coefficient of $x^{n-k}$.
• Avoid Brute Force: Never attempt to calculate each term in a long sum of binomial coefficients unless explicitly directed or if no identity applies. Always seek a pattern and a relevant identity first.

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Summary and Key Takeaway

This problem beautifully demonstrates the power of binomial identities in simplifying complex-looking sums. The sum of the squares of binomial coefficients, $\sum_{k=0}^{n} \left( \binom{n}{k} \right)^2$, always simplifies to $\binom{2n}{n}$. For the given problem, with $n=20$, the sum directly evaluates to ${}^{40}{C_{20}}$. A solid understanding of these identities and their derivations is crucial for efficient problem-solving in combinatorics.