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:

Step 1: Expand the given expression using binomial theorem

We use the binomial theorem to expand each term. The binomial theorem states that $(1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \dots$. We apply this to $(1+x)^p$ and $(1-x)^q$, noting that for $(1-x)^q$, the terms alternate in sign.

Step 2: Multiply the expansions and find coefficients of $x$ and $x^2$

Now we multiply the two expanded series. To find the coefficient of $x$, we multiply the constant term from the first expansion by the $x$ term from the second, and the $x$ term from the first by the constant term from the second. Similarly, for the coefficient of $x^2$, we consider all combinations that result in an $x^2$ term: constant with $x^2$, $x$ with $x$, and $x^2$ with constant.

Step 3: Formulate equations from given coefficients

We are given that the coefficient of $x$ is 1 and the coefficient of $x^2$ is -2. We set up two equations based on these given values and the coefficients we derived in the previous step.

Step 4: Simplify and solve the system of equations

We simplify Equation 2 by multiplying by 2 and rearranging terms. We recognize that $p^2 - 2pq + q^2$ is $(p-q)^2$. Substituting $p-q=1$ from Equation 1, we can solve for $p+q$.

Step 5: Solve for $p$ and $q$

Now we have a simple system of two linear equations with two variables ($p$ and $q$). We can add the two equations to find $p$, and subtract them to find $q$.

Step 6: Calculate $p^2 + q^2$

Finally, we substitute the values of $p$ and $q$ we found into the expression $p^2 + q^2$ to get the final answer.

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