Key Concept: Binomial Theorem Expansion

The problem requires us to work with the coefficients of terms in the expansion of a product of two binomial expressions. The fundamental tool for this is the Binomial Theorem, which states that for any non-negative integer $n$, the expansion of $(a+b)^n$ is given by: $(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots + \binom{n}{n}b^n$ For expressions of the form $(1+x)^n$, the expansion simplifies to: $(1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots$ where $\binom{n}{0} = 1$, $\binom{n}{1} = n$, and $\binom{n}{2} = \frac{n(n-1)}{2}$.

When dealing with a product of two such expansions, say $P(x) = (1+x)^p (1-x)^q$, to find the coefficient of a specific power of $x$ (e.g., $x^1$ or $x^2$), we need to consider all possible combinations of terms from each individual expansion that, when multiplied, result in that specific power of $x$.

Step 1: Expanding the Binomial Terms

First, let's write down the expansions of $(1+x)^p$ and $(1-x)^q$ up to the $x^2$ term. We only need these terms because the problem asks for coefficients of $x$ and $x^2$.

For $(1+x)^p$: $(1+x)^p = \binom{p}{0} + \binom{p}{1}x + \binom{p}{2}x^2 + \dots$ Substituting the binomial coefficients: $(1+x)^p = 1 + px + \frac{p(p-1)}{2}x^2 + \dots \quad (*)$

For $(1-x)^q$: We substitute $-x$ for $x$ in the general expansion formula $(1+y)^q$. $(1-x)^q = \binom{q}{0} + \binom{q}{1}(-x) + \binom{q}{2}(-x)^2 + \dots$ Substituting the binomial coefficients and simplifying the powers of $-x$: $(1-x)^q = 1 - qx + \frac{q(q-1)}{2}x^2 + \dots \quad (**)$

Step 2: Finding the Coefficient of $x$

Now, we consider the product of the two expansions: $(1+x)^p (1-x)^q = \left(1 + px + \frac{p(p-1)}{2}x^2 + \dots\right) \left(1 - qx + \frac{q(q-1)}{2}x^2 + \dots\right)$ To find the coefficient of $x^1$ (the term with $x$), we look for combinations of terms from the two series whose powers of $x$ sum to 1:

1. The constant term from $(1+x)^p$ multiplied by the $x$ term from $(1-x)^q$: $1 \cdot (-qx)$
2. The $x$ term from $(1+x)^p$ multiplied by the constant term from $(1-x)^q$: $px \cdot 1$

Combining these, the term containing $x$ in the product is: $(1)(-qx) + (px)(1) = -qx + px = (p-q)x$ The problem states that the coefficient of $x$ is $1$. Therefore, we can set up our first equation: $p-q = 1 \quad \text{(Equation 1)}$

Step 3: Finding the Coefficient of $x^2$

Next, we find the coefficient of $x^2$ in the product. This requires considering all combinations of terms whose powers of $x$ sum to 2:

1. The constant term from $(1+x)^p$ multiplied by the $x^2$ term from $(1-x)^q$: $1 \cdot \left(\frac{q(q-1)}{2}x^2\right)$
2. The $x$ term from $(1+x)^p$ multiplied by the $x$ term from $(1-x)^q$: $(px) \cdot (-qx)$
3. The $x^2$ term from $(1+x)^p$ multiplied by the constant term from $(1-x)^q$: $\left(\frac{p(p-1)}{2}x^2\right) \cdot 1$

Combining these, the terms containing $x^2$ in the product are: $(1)\left(\frac{q(q-1)}{2}x^2\right) + (px)(-qx) + \left(\frac{p(p-1)}{2}x^2\right)(1)$ $= \left(\frac{q(q-1)}{2} - pq + \frac{p(p-1)}{2}\right)x^2$ The problem states that the coefficient of $x^2$ is $-2$. So, we set up our second equation: $\frac{q(q-1)}{2} - pq + \frac{p(p-1)}{2} = -2$ To simplify, multiply the entire equation by 2 to eliminate the denominators: $q(q-1) - 2pq + p(p-1) = -4$ Expand the terms: $q^2 - q - 2pq + p^2 - p = -4$ Rearrange the terms to group $p^2$, $-2pq$, and $q^2$ together, and $p$ and $q$ separately: $(p^2 - 2pq + q^2) - (p+q) = -4$ We recognize the term in the parenthesis as the expansion of $(p-q)^2$: $(p-q)^2 - (p+q) = -4 \quad \text{(Equation 2')}$

Step 4: Solving the System of Equations

Now we have a system of two equations with two variables, $p$ and $q$:

1. $p-q = 1$
2. $(p-q)^2 - (p+q) = -4$

Substitute Equation 1 into Equation 2': $(1)^2 - (p+q) = -4$ $1 - (p+q) = -4$ To solve for $p+q$, add $(p+q)$ to both sides and add 4 to both sides: $1+4 = p+q$ $p+q = 5 \quad \text{(Equation 2)}$

Now we have a simpler system of linear equations: $p-q = 1$ $p+q = 5$ Add the two equations together: $(p-q) + (p+q) = 1 + 5$ $2p = 6$ $p = 3$ Substitute the value of $p=3$ into Equation 1: $3-q = 1$ $q = 3-1$ $q = 2$ So, we found that $p=3$ and $q=2$.

Step 5: Calculating the Final Value

The problem asks for the value of $p^2+q^2$. $p^2+q^2 = (3)^2 + (2)^2$ $p^2+q^2 = 9 + 4$ $p^2+q^2 = 13$

Tips and Common Mistakes

• Sign Errors: Be extremely careful with signs, especially when expanding $(1-x)^q$. A common mistake is to forget the alternating signs or to incorrectly square $-x$.
• Missing Combinations: When finding coefficients in a product of series, ensure you consider ALL combinations of terms that lead to the desired power of $x$. For $x^2$, this involves three distinct pairs of terms.
• Algebraic Manipulation: Pay attention to algebraic simplification. Recognizing expressions like $(p^2 - 2pq + q^2)$ as $(p-q)^2$ can significantly simplify the process.
• Understanding Binomial Coefficients: Remember the basic definitions: $\binom{n}{0}=1$, $\binom{n}{1}=n$, $\binom{n}{2}=\frac{n(n-1)}{2}$.

Summary and Key Takeaway

This problem is a classic application of the Binomial Theorem combined with careful algebraic manipulation. The key steps involve:

1. Expanding each binomial expression up to the required power of $x$.
2. Systematically identifying and summing the coefficients of $x^1$ and $x^2$ from the product of the expansions.
3. Setting up a system of linear equations based on the given coefficients.
4. Solving the system to find the values of $p$ and $q$.
5. Finally, calculating the expression $p^2+q^2$. Mastering the technique of finding coefficients in a product of series is crucial for such problems.