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 :

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 \(\mathrm{p}^{2}+\mathrm{q}^{2}\) is equal to :

(1) 8

(2) 18

(3) 13

(4) 20

Answer 1

Correct option is (3) 13 

\((1+\mathrm{x})^{\mathrm{p}}(1-\mathrm{x})^{q}=\left({ }^{\mathrm{p}} \mathrm{C}_{0}+{ }^{\mathrm{p}} \mathrm{C}_{1} \mathrm{x}+{ }^{\mathrm{p}} \mathrm{C}_{2} \mathrm{x}^{2}+\ldots\right)\left({ }^{q} \mathrm{C}_{0}-{ }^{q} \mathrm{C}_{1} \mathrm{x}+{ }^{q} \mathrm{C}_{2} \mathrm{x}^{2}+\ldots\right)\)

coff of \(\mathrm{x} \equiv{ }^{\mathrm{p}} \mathrm{C}_{0}{ }^{\mathrm{q}} \mathrm{C}_{1}-{ }^{\mathrm{p}} \mathrm{C}_{1}{ }^{\mathrm{q}} \mathrm{C}_{0}=1\)

p - q = 1

coff of \(x^{2} \equiv{ }^{\mathrm{p}} \mathrm{C}_{0}{ }^{\mathrm{q}} \mathrm{C}_{2}-{ }^{\mathrm{p}} \mathrm{C}_{1}{ }^{\mathrm{q}} \mathrm{C}_{1}+{ }^{\mathrm{p}} \mathrm{C}_{2}{ }^{\mathrm{q}} \mathrm{C}_{0}=-2\)

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

\(q^{2}-q-2 p q+p^{2}-p=-4\)

\((p-q)^{2}-(p+q)=-4\)

p + q = 5

\(\mathrm{p}=3\)

\(\mathrm{q}=2\)

so \(\mathrm{p}^{2}+\mathrm{q}^{2}=13\)