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

This problem is solved by analyzing the coefficients in the expansion of \( (1 + x)^p (1 - x)^q \).

Step 1: Expansion Analysis

The binomial expression is \( (1 + x)^p (1 - x)^q \). The expansion of \( (1 + x)^p \) is:

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

The expansion of \( (1 - x)^q \) is:

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

Multiplying these series involves combining terms with the same power of \( x \).

Step 2: Coefficient of \( x \)

The coefficient of \( x \) is derived from terms producing \( x^1 \):

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

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

\(p - q = 1\) (Equation 1)

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

The coefficient of \( x^2 \) is determined by:

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

This simplifies to:

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

Step 4: Solving the System

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

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

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

After expansion and simplification:

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

With \( q = 2 \), we find \( p = q + 1 = 3 \).

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

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

The value of \( p^2 + q^2 \) is 13.