If (1 + x)ⁿ = C₀ + C₁x + C₂x² + .... + Cⁿ xⁿ then show that the sum of the products of the Ci's take

Q: Wait, why do we use the square of the sum? Is there another way?

While you could try brute force, the square of the sum is the most elegant path. It maps perfectly to the structure of the problem. By squaring the sum, you naturally generate all possible cross-products $C_iC_j$, which is exactly what the problem asks for.

Q: How do I know $\sum C_i^2 = \binom{2n}{n}$?

It is a classic result derived from the identity $(1+x)^n(x+1)^n = (1+x)^{2n}$. If you expand both sides and compare the coefficient of $x^n$, you will find that $\sum C_i^2$ is indeed equal to the coefficient of $x^n$ in $(1+x)^{2n}$, which is $\binom{2n}{n}$.

Q: Why is the condition $i < j$ important?

The condition $i < j$ ensures that we are counting each pair of distinct coefficients exactly once. In the expansion of the square, we get $C_iC_j$ for all $i, j$. Since $C_iC_j = C_jC_i$, the sum $\sum_{i \neq j} C_iC_j$ is twice the sum $\sum_{i < j} C_iC_j$. This is why the factor of $2$ appears in our identity!

Question

If $(1 + x)^n = C_0 + C_1x + C_2x^2 + .... + C_nx^n$ then show that the sum of the products of the $C_i$'s taken two at a time, represented by $\sum_{0 \le i < j \le n} C_iC_j$ is equal to $2^{2n-1} - \frac{(2n)!}{2(n!)^2}$

Sigma Percentile · Accepted Answer

### The Beauty of Binomial Symmetry

Welcome, fellow traveler of the mathematical landscape! Today, we are going to peel back the layers of a classic Binomial Theorem problem.

We are given the expansion $(1 + x)^n = C_0 + C_1x + C_2x^2 + \dots + C_nx^n$, and our mission is to find the sum of the products of these coefficients taken two at a time: $\sum_{0 \le i < j \le n} C_i C_j$.

This might look like a daunting task, but it is actually a beautiful exercise in symmetry and algebraic structure.

### The Grid Analogy

To understand this sum of products, let us visualize an algebraic identity. Imagine a square grid where we multiply the sum of coefficients by itself: $(\sum_{i=0}^n C_i)^2$.

If you were to expand this, you would get a collection of terms. The diagonal elements of this grid are the squares of each coefficient, $C_i^2$.

The off-diagonal elements are the products $C_iC_j$. Because the grid is symmetric, the upper triangle (where $i < j$) is identical to the lower triangle (where $i > j$).

Thus, the total sum can be written as:

$$(\sum_{i=0}^n C_i)^2 = \sum_{i=0}^n C_i^2 + 2 \sum_{0 \le i < j \le n} C_i C_j$$

This identity is our master key.

### Unlocking the Components

Now, we need to find the values of the two main components: the total sum of coefficients and the sum of their squares.

First, finding $\sum C_i$ is straightforward. By substituting $x = 1$ into our original expansion $(1+x)^n$, we get $(1+1)^n = C_0 + C_1 + \dots + C_n$, which simplifies to $2^n$.

Squaring this gives us $2^{2n}$.

Next, we tackle the sum of squares, $\sum C_i^2$. This is a famous result in combinatorics.

The sum of the squares of binomial coefficients is equal to $\binom{2n}{n}$, which in factorial form is:

$$\frac{(2n)!}{(n!)^2}$$

This represents the diagonal of our grid.

### The Final Synthesis

We are now ready to assemble the pieces. Substituting our values into the identity, we have:

$$2^{2n} = \frac{(2n)!}{(n!)^2} + 2 \sum_{0 \le i < j \le n} C_i C_j$$

Our goal is to isolate the double summation. We subtract the factorial term from both sides:

$$2 \sum_{0 \le i < j \le n} C_i C_j = 2^{2n} - \frac{(2n)!}{(n!)^2}$$

Finally, we divide the entire equation by $2$. Dividing $2^{2n}$ by $2$ gives us $2^{2n-1}$, and dividing the factorial term by $2$ gives us $\frac{(2n)!}{2(n!)^2}$.

And there it is! The sum of products taken two at a time is:

$$\mathbf{2^{2n-1} - \frac{(2n)!}{2(n!)^2}}$$

We have successfully navigated the complexity and arrived at the elegant solution. Keep this identity in your toolkit; it is a powerful weapon for any JEE aspirant!

Visualized Solution

The Binomial Expansion & Goal

The Square of a Sum Identity

Sum of Coefficients $\sum C_{i}$

Sum of Squares $\sum C_{i}^{2}$

Substituting into the Identity

Isolating the Product Term

The Final Result

Key Takeaways

The Sigma Insight: Properties of Binomial Coefficients

The Beauty of Binomial Symmetry

The Grid Analogy

Unlocking the Components

The Final Synthesis

Similar Questions

Given that $C_{1} + 2 C_{2} x + 3 C_{3} x^{2} + .... + 2 n C_{2 n} x^{2 n - 1} = 2 n (1 + x)^{2 n - 1}$ where $C_{r} = \frac{( 2 n )!}{r ! ( 2 n - r )!}, r = 0, 1, 2, ...., 2 n$ . Prove that $C_{1}^{2} - 2 C_{2}^{2} + 3 C_{3}^{2} - .... - 2 n C_{2 n}^{2} = (- 1)^{n} n^{n} C_{n}$ .

Let $n$ be a positive integer and $(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + .... + a_{2 n} x^{2 n}$ . Show that $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - .... + a_{2 n}^{2} = a_{n}$ .

Given $s_{n} = 1 + q + q^{2} + .... + q^{n}$ ; $S_{n} = 1 + \frac{q + 1}{2} + (\frac{q + 1}{2})^{2} + .... + (\frac{q + 1}{2})^{n}, q \neq = 1$ Prove that $^{n + 1} C_{1} +^{n + 1} C_{2} s_{1} +^{n + 1} C_{3} s_{2} + .... +^{n + 1} C_{n + 1} s_{n} = 2^{n} S_{n}$

For any positive integer $m, n$ (with $n \geq m$ ), let $(m n) =^{n} C_{m}$ . Prove that $(m n) + (m n - 1) + (m n - 2) + .... + (m m) = (m + 1 n + 1)$ . Hence or otherwise, prove that $(m n) + 2 (m n - 1) + 3 (m n - 2) + .... + (n - m + 1) (m m) = (m + 2 n + 2)$ .

Prove that $2^{k} (0 n) (k n) - 2^{k - 1} (1 n) (k - 1 n - 1) + 2^{k - 2} (2 n) (k - 2 n - 2) - .... + (- 1)^{k} (k n) (0 n - k) = (k n)$

Using mathematical induction, prove that $^{m} C_{0}^{n} C_{k} +^{m} C_{1}^{n} C_{k - 1} + .... +^{m} C_{k}^{n} C_{0} =^{(m + n)} C_{k}$ , where $m, n, k$ are positive integers, and $^{p} C_{q} = 0$ for $p < q$ .

If $\sum_{r = 0}^{2 n} a_{r} (x - 2)^{r} = \sum_{r = 0}^{2 n} b_{r} (x - 3)^{r}$ and $a_{k} = 1$ for all $k \geq n$ , then show that $b_{n} =^{2 n + 1} C_{n + 1}$ .

Prove that $C_{0} - 2^{2} C_{1} + 3^{2} C_{2} - .... + (- 1)^{n} (n + 1)^{2} C_{n} = 0, n > 2$ , where $C_{r} =^{n} C_{r}$ .

Statement-1: $\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1}$ . Statement-2: $\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1}$ .

$\sum_{i, j = 0 i \neq = j}^{n}^{n} C_{i}^{n} C_{j}$ is equal to

Visualized Solution

The Binomial Expansion & Goal

The Square of a Sum Identity

Sum of Coefficients .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑Ci​

Sum of Squares .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑Ci2​