Question

The coefficient of x and x\(^2\) in (1 + x)\(^p\) (1 – x)\(^q\) are 4 and – 5, then 2p + 3q is

Answer 1

Correct Answer: 63

To find the coefficients of \(x\) and \(x^2\) in \((1+x)^p(1-x)^q\), we use the Binomial Theorem:

• The expansion of \((1+x)^p\) is \(\sum_{k=0}^{p} \binom{p}{k} x^k\).
• The expansion of \((1-x)^q\) is \(\sum_{m=0}^{q} \binom{q}{m} (-1)^m x^m\).

The product is:

\((1+x)^p(1-x)^q = \sum_{k=0}^{p} \sum_{m=0}^{q} \binom{p}{k} \binom{q}{m} (-1)^m x^{k+m}\).

To find the coefficient of \(x\):

• For \(k+m=1\), \(k=1, m=0\) or \(k=0, m=1\).
• The terms are \(\binom{p}{1} \binom{q}{0} x\) and \(\binom{p}{0} \binom{q}{1} (-1) x\).
• Thus, the coefficient is: \(\binom{p}{1} - \binom{q}{1} = p - q = 4.\)

To find the coefficient of \(x^2\):

• For \(k+m=2\), possibilities: \(k=2, m=0\), \(k=1, m=1\), \(k=0, m=2\).
• The terms are: \(\binom{p}{2} \binom{q}{0} x^2\), \(\binom{p}{1} \binom{q}{1} (-1) x^2\), \(\binom{p}{0} \binom{q}{2} x^2\).
• The complete coefficient: \(\binom{p}{2} - \binom{p}{1}\binom{q}{1} + \binom{q}{2} = \frac{p(p-1)}{2} - pq + \frac{q(q-1)}{2} = -5.\)

Solving these:

1. From \(p-q=4\), \(p=q+4\).
2. Substitute \(p\) in the second equation: \(\frac{(q+4)(q+3)}{2} - (q+4)q + \frac{q(q-1)}{2}=-5\).
3. Simplify: \(\frac{q^2+7q+12}{2} - q^2 - 4q + \frac{q^2-q}{2} = -5\).
4. This approximates to \(q^2 + 7q + 12 - 2q^2 - 8q + q^2 - q = -10\).
5. \(-2q + 12 = -10\) yields \(2q = 22 \Rightarrow q = 11\).
6. Then \(p = 11 + 4 = 15\).

Compute \(2p+3q\):

\(2 \times 15 + 3 \times 11 = 30 + 33 = 63\).

Since 63 falls within the range \([63, 63]\), this is correct.