Σ (²⁰Ck)² from k=0 to 20 is equal to : (1) ⁴⁰C₂₁ (2) ⁴⁰C₁₉ (3) ⁴⁰C₂₀ (4) ⁴¹C₂₀

Q: Wait, why can we just replace ${}^{20}C_k$ with ${}^{20}C_{20-k}$? Does that change the value of the sum?

That is a brilliant observation! Because the binomial coefficient is symmetric, ${}^{20}C_k$ is exactly equal to ${}^{20}C_{20-k}$. When we sum from $k=0$ to $20$, we are just adding the same terms in a different order. It is like saying $1+2+3$ is the same as $3+2+1$. The sum remains identical, but the new form reveals a hidden structure!

Q: Why do we need to consider two groups of 20? The problem only mentions one group of 20.

Think of it as a 'combinatorial trick.' We are creating a mental model to solve a difficult algebraic problem. By splitting the 40 items into two groups of 20, we can use the logic of 'choosing $k$ from the first and $20-k$ from the second' to represent the total selection of 20 items. It is a powerful way to visualize the identity.

Q: Is there a way to solve this without using the 'two groups' logic?

Absolutely! You can use the Generating Functions method. By expanding $(1+x)^{20} \cdot (1+x)^{20} = (1+x)^{40}$, you can look for the coefficient of $x^{20}$ on both sides. It is a more algebraic approach that leads to the exact same beautiful result.

Q: Why is the answer ${}^{40}C_{20}$ and not something like ${}^{40}C_{21}$?

Great question! The sum $\sum_{k=0}^{20} {}^{20}C_k \cdot {}^{20}C_{20-k}$ represents the total number of ways to pick exactly 20 items from a pool of 40. Since we are picking 20 items total, the result must be ${}^{40}C_{20}$. If we were picking a different number, the index would change accordingly.

Q: Does this identity work for any $n$, or just for $n=20$?

It is a universal truth! The identity $\sum_{k=0}^{n} ({}^nC_k)^2 = {}^{2n}C_n$ holds for any positive integer $n$. You have just discovered a fundamental property of binomial coefficients that will serve you well in many JEE problems.

Question

$\sum_{k=0}^{20} ({}^{20}C_k)^2$ is equal to :

Options:

(A) ${}^{40}C_{21}$

(B) ${}^{40}C_{19}$

(C) ${}^{40}C_{20}$

(D) ${}^{41}C_{20}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the road to JEE excellence. Today, we are not just solving a summation; we are peeling back the layers of a beautiful mathematical identity.

You are looking at the expression:

$$S = \sum_{k=0}^{20} ({}^{20}C_k)^2$$

At first glance, it looks like a daunting list of squares: $({}^{20}C_0)^2 + ({}^{20}C_1)^2 + \dots + ({}^{20}C_{20})^2$. But I want you to pause and breathe. Mathematics is rarely about brute force; it is about finding the hidden symmetry that makes the complex simple.

### The Symmetry Spark

Recall the fundamental symmetry of binomial coefficients: ${}^{n}C_{k} = {}^{n}C_{n-k}$. This is the key that unlocks the door.

If we take our squared term $({}^{20}C_k)^2$, we can write it as the product of two distinct choices: ${}^{20}C_k \cdot {}^{20}C_k$. Now, apply the symmetry property to the second factor. We replace ${}^{20}C_k$ with ${}^{20}C_{20-k}$.

Suddenly, our sum transforms into something much more suggestive:

$$S = \sum_{k=0}^{20} {}^{20}C_k \cdot {}^{20}C_{20-k}$$

### The Combinatorial Story

Imagine you have two baskets. Basket A contains 20 distinct red balls, and Basket B contains 20 distinct blue balls. You have a total of 40 balls.

Suppose you want to choose exactly 20 balls in total from these 40. How could you do it? You could pick $k$ balls from Basket A and $20-k$ balls from Basket B.

For a fixed $k$, the number of ways to do this is ${}^{20}C_k \cdot {}^{20}C_{20-k}$. If you sum this over all possible values of $k$ (from 0 to 20), you are accounting for every possible way to pick 20 balls from the combined pool of 40.

This is the physical soul of **Vandermonde's Identity**. The total number of ways to pick 20 items from 40 is simply ${}^{40}C_{20}$.

### The Algebraic Mirror

If the combinatorial story feels too abstract, let us look at the algebra. Consider the binomial expansion of $(1+x)^{20}$:

$$(1+x)^{20} = \sum_{k=0}^{20} {}^{20}C_k x^k$$

Now, multiply this by itself: $(1+x)^{20} \cdot (1+x)^{20} = (1+x)^{40}$.

On the left-hand side, when we multiply the two summations, the coefficient of $x^{20}$ is formed by picking $x^k$ from the first expansion and $x^{20-k}$ from the second. This gives us exactly our sum: $\sum_{k=0}^{20} {}^{20}C_k \cdot {}^{20}C_{20-k}$.

On the right-hand side, the coefficient of $x^{20}$ in $(1+x)^{40}$ is simply ${}^{40}C_{20}$.

### The Final Revelation

Whether you view it through the lens of selecting balls from baskets or through the algebraic expansion of polynomials, the result is the same. We have arrived at the conclusion that:

$$\sum_{k=0}^{20} ({}^{20}C_k)^2 = {}^{40}C_{20}$$

This is not just an answer to a question; it is a testament to the power of perspective. When you face a problem that seems like a mountain of calculation, stop. Look for the symmetry. Look for the story. The math will always guide you home.

Animated Solution for Mathematics - Binomial Theorem: $\sum_{k = 0}^{20} (^{20} C_{k})^{2}$ is equal to :

Select Answer:

Visualized Solution

Analyze the Summation

Apply the Symmetry Property

The Combinatorial Perspective - Setup

Selecting the Items

Summing Over All Possibilities

The Total Selection (Vandermonde's Identity)

Alternative Method: Generating Functions

Multiplying the Expansions

Extracting the Coefficient

Conclusion

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Symmetry Spark

The Combinatorial Story

The Algebraic Mirror

The Final Revelation

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Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑k=020​(20Ck​)2 is equal to :