If in the expansion of (1+x)^p(1-x)^q the coefficients of x and x^2 are 1 and -2, respectively, then

Q: Wait, why can't we just use the general term formula $T_{r+1}$ for the whole expression?

That is a brilliant question! The general term formula $T_{r+1} = \binom{n}{r}x^r$ works perfectly for a single binomial expansion like $(1+x)^p$. However, here we have a product of two different binomials. There is no single 'general term' for the product. We must expand both series individually and then perform the multiplication, collecting terms of the same degree. It is a bit more manual, but it is the only way to handle the product correctly!

Q: Why is the coefficient of $x^2$ so messy? Is there a mistake in my calculation?

It is not messy; it is just rich with information! When you multiply two polynomials, the coefficient of $x^2$ is formed by the sum of all combinations that result in $x^2$. You have the constant from the first times the $x^2$ from the second, the $x$ from the first times the $x$ from the second, and the $x^2$ from the first times the constant from the second. It feels like a lot, but once you write it down, you will see it simplifies beautifully into an algebraic identity. Trust the process!

Q: Can we solve for $p$ and $q$ directly without using the identity $(p-q)^2$?

You absolutely can! You could expand the terms, get a system of equations, and solve for $p$ and $q$ using substitution or elimination. However, in JEE Advanced, time is your most precious resource. Recognizing that $p^2 + q^2 - 2pq$ is $(p-q)^2$ saves you precious minutes and reduces the chance of arithmetic errors. Always look for those hidden algebraic structures!

Question

If in the expansion of $(1+x)^{p}(1-x)^{q}$ the coefficients of $x$ and $x^{2}$ are 1 and -2, respectively, then $p^{2}+q^{2}$ is equal to:

Options:

(A) 18

(B) 13

(C) 8

(D) 20

Sigma Percentile · Accepted Answer

### The Art of the Binomial Dance

Welcome, future engineer. Today, we are not just solving a problem; we are performing a dance of algebra. We are looking at the product of two binomial expansions: $(1+x)^p$ and $(1-x)^q$.

At first glance, this might look like a daunting task—a product of two infinite series. But in the world of JEE Advanced, we are masters of simplification. We only care about the coefficients of $x$ and $x^2$. Anything beyond that is just noise.

Let us strip away the complexity and find the core.

### Phase 1: The Expansion

First, we invoke the Binomial Theorem. For any power $n$, we know that $(1+x)^n = 1 + nx + \frac{n(n-1)}{2}x^2 + \dots$.

Applying this to our two terms, we get:

$$(1+x)^p = 1 + px + \frac{p(p-1)}{2}x^2 + \dots$$

$$(1-x)^q = 1 - qx + \frac{q(q-1)}{2}x^2 - \dots$$

Notice the alternating sign in the second expansion? That is the trap where many students stumble. Keep your signs sharp. We have our two series; now, we must multiply them.

### Phase 2: The Hunt for Coefficients

To find the coefficient of $x$, we look for ways to combine terms from these two series to get a power of $x^1$. We can take the constant $1$ from the first series and multiply it by $-qx$ from the second, or take $px$ from the first and multiply it by the constant $1$ from the second.

Adding these gives us $(p-q)x$. The problem states this coefficient is $1$.

Thus, our first pillar of truth is established: 
$$p - q = 1$$

Now, we address the $x^2$ term. This is where the dance gets intricate. To get an $x^2$ term, we have three paths:

1. The constant $1$ from the first series multiplied by $\frac{q(q-1)}{2}x^2$ from the second.
2. The $px$ from the first series multiplied by $-qx$ from the second, which gives $-pqx^2$.
3. The $\frac{p(p-1)}{2}x^2$ from the first series multiplied by the constant $1$ from the second.

Summing these, we get the coefficient: 
$$\frac{q(q-1)}{2} - pq + \frac{p(p-1)}{2} = -2$$

This looks like a mess, but do not panic. Multiply by $2$ to clear the fractions: 
$$q(q-1) - 2pq + p(p-1) = -4$$

### Phase 3: The Algebraic Elegance

Expand the brackets: 
$$q^2 - q - 2pq + p^2 - p = -4$$

Now, group the terms: 
$$(p^2 + q^2 - 2pq) - (p + q) = -4$$

Do you see it? The term in the parenthesis is the perfect square $(p-q)^2$. We already know $p-q = 1$. Substituting this, we get:
$$1^2 - (p+q) = -4$$

This simplifies to:
$$p+q = 5$$

### The Final Victory

We now have a simple system: $p-q=1$ and $p+q=5$. Adding them gives $2p=6$, so $p=3$. Subtracting them gives $2q=4$, so $q=2$.

The question asks for $p^2 + q^2$. That is:
$$3^2 + 2^2 = 9 + 4 = 13$$

We have arrived at the destination. Remember, in JEE, it is never just about the answer; it is about the elegance of the path you take to get there. The final result is **13**.

Animated Solution for Mathematics - Binomial Theorem: If in the expansion of $(1 + x)^{p} (1 - x)^{q}$ the coefficients of $x$ and $x^{2}$ are 1 and -2, respectively, then $p^{2} + q^{2}$ is equal to:

Select Answer:

Visualized Solution

Problem Setup

Binomial Expansions

Extracting Coefficient of $x$

Forming Equation 1

Extracting Coefficient of $x^{2}$

Forming Equation 2

Simplifying the Equation

Expanding and Grouping

Applying Algebraic Identity

Substituting Known Value

Finding $p + q$

Solving for $p$ and $q$

Final Calculation

The Sigma Insight: Binomial Expansion for Positive Integral Index

The Art of the Binomial Dance

Phase 1

Phase 2

Phase 3

The Final Victory

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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; } If in the expansion of (1+x)p(1−x)q the coefficients of x and x2 are 1 and -2, respectively, then p2+q2 is equal to: