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

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 \( p^2 + q^2 = 13 \).