Given that C₁ + 2C₂x + 3C₃x² + .... + 2nC₂ⁿ x²ⁿ⁻¹ = 2n(1 + x)²ⁿ⁻¹ where Cr = (2n)! / (r!(2n - r)!),

Q: Wait, why did we choose $(1-x)^{2n}$ as our auxiliary series? Couldn't we have used something else?

That is a brilliant question! We chose $(1-x)^{2n}$ for two critical reasons. First, the alternating signs in your target expression ($-2C_2^2, -4C_4^2$, etc.) demand a series with alternating signs, which $(1-x)^{2n}$ provides via the $(-1)^k$ term. Second, the symmetry property $C_r = C_{2n-r}$ is the engine that allows us to transform the product of coefficients into the square of a coefficient. Without this specific choice, the powers of $x$ wouldn't align to give us the target sum.

Q: Why are we looking for the coefficient of $x^{2n-1}$ specifically? How did we know that was the right power?

Think about the general term of the product. We are multiplying a term with $x^{r-1}$ from the first series by a term with $x^k$ from the second. To get a fixed power, say $x^P$, we need $(r-1) + k = P$. By choosing $P = 2n-1$, we force $k = 2n-r$. Because of the symmetry property $C_k = C_{2n-r} = C_r$, this choice of $P$ perfectly transforms the product of coefficients into $C_r \times C_r$, which is $C_r^2$. It is a deliberate, strategic choice to make the algebra collapse into the form we need.

Q: I got stuck on the RHS simplification. Why does the first term $2n(1-x^2)^{2n-1}$ contribute zero to the coefficient of $x^{2n-1}$?

Don't be discouraged! Look at the structure of $(1-x^2)^{2n-1}$. Every term in its expansion is of the form $\binom{2n-1}{k} (-x^2)^k = \binom{2n-1}{k} (-1)^k x^{2k}$. Notice that the power of $x$ is always even ($2k$). Since $2n-1$ is an odd number (for any integer $n \ge 1$), it is impossible for an even power to equal an odd power. Therefore, the coefficient of $x^{2n-1}$ in that specific part of the expression must be zero. It's a beautiful example of parity saving us from tedious calculation!

Question

Given that $C_1 + 2C_2x + 3C_3x^2 + .... + 2nC_{2n}x^{2n-1} = 2n(1 + x)^{2n-1}$ where $C_r = \frac{(2n)!}{r!(2n - r)!}, r = 0, 1, 2, ...., 2n$. Prove that $C_1^2 - 2C_2^2 + 3C_3^2 - .... - 2nC_{2n}^2 = (-1)^n n ^nC_n$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are witnessing a beautiful, choreographed dance of binomial coefficients.

When you first look at an expression like $C_1^2 - 2C_2^2 + 3C_3^2 - \dots - 2nC_{2n}^2$, it might seem intimidating. It looks like a chaotic mess of squares and alternating signs.

In the world of JEE Advanced, whenever you see squares of binomial coefficients, your intuition should immediately light up. This is not a problem to be brute-forced; it is a problem to be solved through the elegant technique of multiplying two series.

### The Strategy: Finding the Partner

We are given the identity:

$$ \sum_{r=1}^{2n} r C_r x^{r-1} = 2n(1+x)^{2n-1} $$

Our goal is to extract the sum of squares. To do this, we need a partner—an auxiliary series. We need a series that, when multiplied by our given one, will produce terms involving $C_r^2$.

We choose the expansion of:

$$ (1-x)^{2n} = \sum_{k=0}^{2n} (-1)^k C_k x^k $$

The alternating signs in our target expression are perfectly mirrored by the $(-1)^k$ in this expansion. Furthermore, we rely on the fundamental symmetry of binomial coefficients: $C_r = C_{2n-r}$. This is our secret weapon, allowing us to bridge the gap between two different coefficients and turn them into a square.

### The Collision of Series

Now, imagine the two series colliding. We multiply the given identity by our auxiliary expansion. On the left-hand side, we have the product of two summations.

On the right-hand side, we have the product of two compact algebraic expressions:

$$ P(x) = 2n(1+x)^{2n-1} 	imes (1-x)^{2n} $$

To extract our target, we look for the coefficient of $x^{2n-1}$ on both sides. When we multiply a term $x^{r-1}$ from the first series by a term $x^{2n-r}$ from the second, the powers add up to $(r-1) + (2n-r) = 2n-1$.

The coefficient of this term becomes:

$$ \sum_{r=1}^{2n} (r C_r) \cdot (-1)^{2n-r} C_{2n-r} $$

Using our symmetry property $C_{2n-r} = C_r$, this simplifies beautifully to $\sum_{r=1}^{2n} (-1)^r r C_r^2$. If you expand this, you will see it is exactly the negative of our target expression.

### The Algebraic Cleanup

Now, we turn our attention to the right-hand side, $P(x) = 2n(1+x)^{2n-1}(1-x)^{2n}$. We can rewrite $(1-x)^{2n}$ as $(1-x)^{2n-1}(1-x)$.

This allows us to group the terms:

$$ P(x) = 2n [(1+x)(1-x)]^{2n-1} (1-x) = 2n(1-x^2)^{2n-1}(1-x) $$

Expanding this gives us $2n(1-x^2)^{2n-1} - 2nx(1-x^2)^{2n-1}$. We need the coefficient of $x^{2n-1}$ here.

The first term, $2n(1-x^2)^{2n-1}$, only contains even powers of $x$, so it contributes nothing to our odd power $2n-1$. The second term, $-2nx(1-x^2)^{2n-1}$, requires us to find the coefficient of $x^{2n-2}$ in $(1-x^2)^{2n-1}$.

Using the general term $\binom{2n-1}{k} (-x^2)^k$, we set $2k = 2n-2$, which gives $k = n-1$. The coefficient is $\binom{2n-1}{n-1} (-1)^{n-1}$. Multiplying by the $-2n$ outside, we get $2n \binom{2n-1}{n-1} (-1)^n$.

### The Final Victory

We equate the two sides. The negative of our target equals $2n \binom{2n-1}{n-1} (-1)^n$. Therefore, our target equals $2n \binom{2n-1}{n-1} (-1)^{n-1}$.

Finally, we simplify the binomial coefficient:

$$ 2n \binom{2n-1}{n-1} = 2n \frac{(2n-1)!}{(n-1)! n!} = n \frac{(2n)!}{n! n!} = n \binom{2n}{n} $$

And there it is! The chaos has collapsed into order. We have proven that the target expression equals **$(-1)^{n-1} n \binom{2n}{n}$**.

Mathematics is not about memorizing formulas; it is about finding the hidden symmetry in the chaos. Keep practicing, and keep falling in love with the process.

Visualized Solution

Analyze the Given Identity

The Auxiliary Expansion

Multiplying the Two Series

Coefficient of $x^{2 n - 1}$ in LHS

Relating to the Target Expression

Simplifying the RHS Product $P (x)$

Coefficient of $x^{2 n - 1}$ in RHS

Calculating the RHS Coefficient

Final Equating and Simplification

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Strategy

The Collision of Series

The Algebraic Cleanup

The Final Victory

Similar Questions

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