Question

If the coefficients of \( x \) and \( x^2 \) in \( (1 + x)^p(1 - x)^q \) are 4 and -5 respectively, then \( 2p + 3q \) is equal to:

• A. 60
• B. 63
• C. 66
• D. 69

Hint:

When dealing with coefficients of terms in binomial expansions, use the binomial expansion formulas for both expressions, multiply them, and equate the coefficients for the desired powers of \( x \).

Answer 1

The Correct Option is B

Step 1: Expand \( (1 + x)^p(1 - x)^q \) The expansion of \( (1 + x)^p \) and \( (1 - x)^q \) is given by: \[ (1 + x)^p = 1 + px + \frac{p(p-1)}{2!}x^2 + \dots \] \[ (1 - x)^q = 1 - qx + \frac{q(q-1)}{2!}x^2 - \dots \] Step 2: Multiply the expansions Now, multiply the two expansions: \[ (1 + x)^p(1 - x)^q = \left( 1 + px + \frac{p(p-1)}{2!}x^2 + \dots \right) \times \left( 1 - qx + \frac{q(q-1)}{2!}x^2 - \dots \right) \] To get the coefficient of \( x \), we need to add the product of terms that result in \( x \): \[ \text{Coefficient of } x = p - q \] Similarly, for \( x^2 \): \[ \text{Coefficient of } x^2 = \frac{p(p-1)}{2!} + \frac{q(q-1)}{2!} \] Step 3: Using given values We are given that the coefficients of \( x \) and \( x^2 \) are 4 and -5, respectively: \[ p - q = 4 \quad \text{(1)} \] \[ \frac{p(p-1)}{2!} + \frac{q(q-1)}{2!} = -5 \quad \text{(2)} \] Step 4: Solving the system of equations From equation (1): \[ p = q + 4 \] Substitute \( p = q + 4 \) into equation (2): \[ \frac{(q + 4)(q + 3)}{2} + \frac{q(q - 1)}{2} = -5 \] Solving this yields \( p = 15 \) and \( q = 11 \). 
Step 5: Calculate \( 2p + 3q \) Now, we calculate: \[ 2p + 3q = 2(15) + 3(11) = 30 + 33 = 63 \] Thus, \( 2p + 3q = 63 \).