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 :

We need to find $$p^2 + q^2$$ given that in $$(1+x)^p(1-x)^q$$ the coefficients of $$x$$ and $$x^2$$ are 1 and -2, respectively.

Expanding each binomial to second order, we have:

$$(1+x)^p \approx 1 + px + \frac{p(p-1)}{2}x^2 + \ldots$$

$$(1-x)^q \approx 1 - qx + \frac{q(q-1)}{2}x^2 + \ldots$$

Multiplying these series, the coefficient of $$x$$ is

$$p - q = 1 \quad\text{(i)}$$

and the coefficient of $$x^2$$ is

$$\frac{p(p-1)}{2} - pq + \frac{q(q-1)}{2} = -2 \quad\text{(ii)}$$

Multiplying equation (ii) by 2 and simplifying gives:

$$p^2 - p - 2pq + q^2 - q = -4,$$

which can be written as

$$(p - q)^2 - (p + q) = -4.$$

Since (i) implies $$(p - q)^2 = 1,$$ substituting into the last equation yields

$$1 - (p + q) = -4,$$

so

$$p + q = 5 \quad\text{(iii)}.$$

Solving the system $$p - q = 1$$ and $$p + q = 5$$ gives $$p = 3$$ and $$q = 2$$.

Finally,

$$p^2 + q^2 = 9 + 4 = 13.$$

The correct answer is Option 2: 13.