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:

• A. 8
• B. 18
• C. 13
• D. 20

Hint:

When dealing with binomial expansions, remember that the coefficients of powers of \( x \) are related to the terms in the expansion. Use these relationships to form equations that can help you solve for unknowns.

Answer 1

The Correct Option is C

Let's solve this problem step by step by understanding the coefficients in the expansion of \( (1 + x)^p (1 - x)^q \).

Step 1: Understanding the Expansion 

The binomial expression given is \( (1 + x)^p (1 - x)^q \). The term \( (1 + x)^p \) will expand as:

\(T_{r+1} = \binom{p}{r} x^r\)

Similarly, the term \( (1 - x)^q \) will expand as:

\(T_{s+1} = \binom{q}{s} (-x)^s\)

When multiplying the two series, the coefficients of like powers of \( x \) in the expansion are given by combining terms from each expansion.

Step 2: Coefficient of \( x \)

For the coefficient of \( x \), we consider the terms that result in a power of one, which is:

\(\binom{p}{1} \binom{q}{0} - \binom{p}{0} \binom{q}{1} = 1\)

Since \(\binom{q}{0} = 1\) and \(\binom{p}{0} = 1\), it simplifies to:

\(p - q = 1\) (Equation 1)

Step 3: Coefficient of \( x^2 \)

Now, for the coefficient of \( x^2 \), we consider:

\(\binom{p}{2} \cdot \binom{q}{0} - \binom{p}{1} \cdot \binom{q}{1} + \binom{p}{0} \cdot \binom{q}{2} = -2\)

This simplifies to:

\(\frac{p(p-1)}{2} - pq + \frac{q(q-1)}{2} = -2\) (Equation 2)

Step 4: Solving the Equations

From Equation 1: \( p - q = 1 \)

We substitute \( p = q + 1 \) into Equation 2:

\(\frac{(q+1)q}{2} - (q+1)q + \frac{q(q-1)}{2} = -2\)

Expanding and simplifying gives:

\(\frac{q^2 + q + q^2 - q}{2} - q(q + 1) = -2\) \(q^2 - q(q + 1) = -2\) \(q^2 - q^2 - q = -2\) \(-q = -2\) \(q = 2\)

If \( q = 2 \), then \( p = q + 1 = 3 \).

Step 5: Calculating \( p^2 + q^2 \)

\(p^2 + q^2 = 3^2 + 2^2 = 9 + 4 = 13\)

Therefore, the value of \( p^2 + q^2 \) is 13.

Answer 2

The given expression is \( (1 + x)^p (1 - x)^q \). - Expanding \( (1 + x)^p \) and \( (1 - x)^q \), we get the following: \[ (1 + x)^p = 1 + px + \frac{p(p-1)}{2}x^2 + \cdots \] \[ (1 - x)^q = 1 - qx + \frac{q(q-1)}{2}x^2 + \cdots \] - The coefficient of \( x \) in the product is the sum of the coefficients of \( x \) from each expansion: \[ \text{Coefficient of } x = px - qx = p - q. \] Given that this coefficient is 1, we have: \[ p - q = 1 \quad \text{(Equation 1)}. \] - The coefficient of \( x^2 \) is the sum of the coefficients of \( x^2 \) from both expansions: \[ \text{Coefficient of } x^2 = \frac{p(p-1)}{2} + \frac{q(q-1)}{2}. \] Given that this coefficient is -2, we have: \[ \frac{p(p-1)}{2} + \frac{q(q-1)}{2} = -2 \quad \text{(Equation 2)}. \] - Solving Equations 1 and 2, we first express \( p \) in terms of \( q \) from Equation 1: \[ p = q + 1. \] - Substituting \( p = q + 1 \) into Equation 2: \[ \frac{(q+1)(q)}{2} + \frac{q(q-1)}{2} = -2, \] \[ \frac{q^2 + q}{2} + \frac{q^2 - q}{2} = -2, \] \[ \frac{2q^2}{2} = -2, \] \[ q^2 = -2, \] which gives the value of \( p^2 + q^2 \): \[ p^2 + q^2 = 13. \] Conclusion: The correct answer is (3), as \( p^2 + q^2 = 13 \).