Key Concepts and Formulas

This problem utilizes the binomial theorem to find coefficients of specific terms in the expansion of a product of two binomial expressions. The binomial expansion of $(1+x)^n$ is given by: $(1+x)^n = \sum_{k=0}^n {}^nC_k x^k = {}^nC_0 + {}^nC_1 x + {}^nC_2 x^2 + \dots + {}^nC_n x^n$ where ${}^nC_k = \frac{n!}{k!(n-k)!}$.

Similarly, the binomial expansion of $(1-x)^n$ is given by: $(1-x)^n = \sum_{k=0}^n {}^nC_k (-x)^k = {}^nC_0 - {}^nC_1 x + {}^nC_2 x^2 - {}^nC_3 x^3 + \dots + (-1)^n {}^nC_n x^n$

To find the coefficient of $x^k$ in the product of two series, say $A(x) = a_0 + a_1 x + a_2 x^2 + \dots$ and $B(x) = b_0 + b_1 x + b_2 x^2 + \dots$, we multiply the series: $A(x)B(x) = (a_0 + a_1 x + a_2 x^2 + \dots)(b_0 + b_1 x + b_2 x^2 + \dots)$ The coefficient of $x^k$ in the product will be the sum of all $a_i b_j$ such that $i+j=k$. For example, the coefficient of $x^1$ is $a_0b_1 + a_1b_0$. The coefficient of $x^2$ is $a_0b_2 + a_1b_1 + a_2b_0$. The coefficient of $x^3$ is $a_0b_3 + a_1b_2 + a_2b_1 + a_3b_0$.

Step-by-step Derivation

Step 1: Expand $(1+x)^p$ and $(1-x)^q$ up to $x^3$ terms.

We begin by writing out the first few terms of the binomial expansions for $(1+x)^p$ and $(1-x)^q$ using the formulas above. This is done to identify the individual terms that will contribute to the coefficients of $x$, $x^2$, and $x^3$ when these two expansions are multiplied.

$(1+x)^p = {}^pC_0 + {}^pC_1 x + {}^pC_2 x^2 + {}^pC_3 x^3 + \dots$ Using the direct formulas for binomial coefficients: ${}^pC_0 = 1$ ${}^pC_1 = p$ ${}^pC_2 = \frac{p(p-1)}{2}$ ${}^pC_3 = \frac{p(p-1)(p-2)}{6}$ So, $(1+x)^p = 1 + px + \frac{p(p-1)}{2}x^2 + \frac{p(p-1)(p-2)}{6}x^3 + \dots \quad (*)$

Similarly for $(1-x)^q$: $(1-x)^q = {}^qC_0 - {}^qC_1 x + {}^qC_2 x^2 - {}^qC_3 x^3 + \dots$ Using the direct formulas for binomial coefficients and incorporating the alternating signs: ${}^qC_0 = 1$ $-{}^qC_1 = -q$ ${}^qC_2 = \frac{q(q-1)}{2}$ $-{}^qC_3 = -\frac{q(q-1)(q-2)}{6}$ So, $(1-x)^q = 1 - qx + \frac{q(q-1)}{2}x^2 - \frac{q(q-1)(q-2)}{6}x^3 + \dots \quad (**)$

Step 2: Determine the coefficient of $x$.

To find the coefficient of $x$ in the product $(1+x)^p (1-x)^q$, we multiply the expansions $(*)$ and $(**)$ and collect all terms that result in $x^1$. The terms that produce $x^1$ are:

• (constant term from $(1+x)^p$) $\times$ ($x$ term from $(1-x)^q$)
• ($x$ term from $(1+x)^p$) $\times$ (constant term from $(1-x)^q$)

From $(*)$ and $(**)$: Coefficient of $x$ in $(1+x)^p (1-x)^q = ({}^pC_0 \times -{}^qC_1) + ({}^pC_1 \times {}^qC_0)$ Substituting the values: $(1 \times -q) + (p \times 1) = p - q$ We are given that the coefficient of $x$ is $-3$. Therefore, we have our first equation: $p - q = -3 \quad (\text{Equation 1})$

Step 3: Determine the coefficient of $x^2$.

Now, we find the coefficient of $x^2$ in the product. The terms that produce $x^2$ are:

• (constant term from $(1+x)^p$) $\times$ ($x^2$ term from $(1-x)^q$)
• ($x$ term from $(1+x)^p$) $\times$ ($x$ term from $(1-x)^q$)
• ($x^2$ term from $(1+x)^p$) $\times$ (constant term from $(1-x)^q$)

From $(*)$ and $(**)$: Coefficient of $x^2$ in $(1+x)^p (1-x)^q = ({}^pC_0 \times {}^qC_2) + ({}^pC_1 \times -{}^qC_1) + ({}^pC_2 \times {}^qC_0)$ Substituting the values: $\left(1 \times \frac{q(q-1)}{2}\right) + (p \times -q) + \left(\frac{p(p-1)}{2} \times 1\right)$ $= \frac{q(q-1)}{2} - pq + \frac{p(p-1)}{2}$ We are given that the coefficient of $x^2$ is $-5$. So, our second equation is: $\frac{q(q-1)}{2} - pq + \frac{p(p-1)}{2} = -5 \quad (\text{Equation 2})$

Step 4: Solve the system of equations for $p$ and $q$.

We have a system of two linear equations with two variables:

1. $p - q = -3$
2. $\frac{q(q-1)}{2} - pq + \frac{p(p-1)}{2} = -5$

From Equation 1, we can express $p$ in terms of $q$: $p = q - 3$ Now, substitute this expression for $p$ into Equation 2. This strategy allows us to reduce Equation 2 to a single variable equation in $q$, making it solvable.

$\frac{q(q-1)}{2} - (q-3)q + \frac{(q-3)((q-3)-1)}{2} = -5$ $\frac{q(q-1)}{2} - q(q-3) + \frac{(q-3)(q-4)}{2} = -5$ To eliminate the denominators, multiply the entire equation by 2: $q(q-1) - 2q(q-3) + (q-3)(q-4) = -10$ Now, expand and simplify the terms: $(q^2 - q) - (2q^2 - 6q) + (q^2 - 4q - 3q + 12) = -10$ $q^2 - q - 2q^2 + 6q + q^2 - 7q + 12 = -10$ Combine like terms: $(q^2 - 2q^2 + q^2) + (-q + 6q - 7q) + 12 = -10$ $0q^2 + (-2q) + 12 = -10$ $-2q + 12 = -10$ Subtract 12 from both sides: $-2q = -10 - 12$ $-2q = -22$ Divide by $-2$: $q = 11$ Now that we have $q=11$, substitute it back into $p = q-3$: $p = 11 - 3$ $p = 8$ We are given the constraint $p, q \le 15$. Our values $p=8$ and $q=11$ satisfy this condition.

Step 5: Calculate the coefficient of $x^3$.

With $p=8$ and $q=11$, we now need to find the coefficient of $x^3$ in $(1+x)^8 (1-x)^{11}$. The terms that produce $x^3$ are:

• (constant term from $(1+x)^p$) $\times$ ($x^3$ term from $(1-x)^q$)
• ($x$ term from $(1+x)^p$) $\times$ ($x^2$ term from $(1-x)^q$)
• ($x^2$ term from $(1+x)^p$) $\times$ ($x$ term from $(1-x)^q$)
• ($x^3$ term from $(1+x)^p$) $\times$ (constant term from $(1-x)^q$)

Using the general form ${}^nC_k$: Coefficient of $x^3 = ({}^pC_0 \times -{}^qC_3) + ({}^pC_1 \times {}^qC_2) + ({}^pC_2 \times -{}^qC_1) + ({}^pC_3 \times {}^qC_0)$ Substitute $p=8$ and $q=11$: $({}^8C_0 \times -{}^{11}C_3) + ({}^8C_1 \times {}^{11}C_2) + ({}^8C_2 \times -{}^{11}C_1) + ({}^8C_3 \times {}^{11}C_0)$ Now, calculate each binomial coefficient:

• ${}^8C_0 = 1$
• ${}^{11}C_3 = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 11 \times 5 \times 3 = 165$
• ${}^8C_1 = 8$
• ${}^{11}C_2 = \frac{11 \times 10}{2 \times 1} = 55$
• ${}^8C_2 = \frac{8 \times 7}{2 \times 1} = 28$
• ${}^{11}C_1 = 11$
• ${}^8C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 8 \times 7 = 56$
• ${}^{11}C_0 = 1$

Substitute these values back into the expression for the coefficient of $x^3$: $(1 \times -165) + (8 \times 55) + (28 \times -11) + (56 \times 1)$ $= -165 + 440 - 308 + 56$ Perform the addition and subtraction: $= (440 + 56) - (165 + 308)$ $= 496 - 473$ $= 23$

Thus, the coefficient of $x^3$ is $23$.

Tips and Common Mistakes to Avoid

1. Sign Errors: Be extremely careful with the alternating signs in the expansion of $(1-x)^q$. A common mistake is to forget the negative signs for odd powers of $x$.
2. Incorrect Binomial Coefficient Calculation: Ensure accurate calculation of ${}^nC_k$. Memorize or have a quick way to derive ${}^nC_0, {}^nC_1, {}^nC_2, {}^nC_3$ as these are frequently used.
3. Algebraic Errors: The most common pitfalls in this problem type are algebraic mistakes when solving the simultaneous equations. Double-check each step of simplification and substitution.
4. Systematic Approach: When finding coefficients in a product of expansions, it's crucial to be systematic. List out all combinations of terms that multiply to give the desired power of $x$ (e.g., constant $\times x^k$, $x^1 \times x^{k-1}$, etc.) to ensure no terms are missed.

Summary and Key Takeaway

This problem effectively tests the understanding of binomial expansions and the ability to find coefficients in the product of two polynomial series. The core idea is to first establish a system of equations for the unknown powers ($p$ and $q$) using the given coefficients. Once these values are found, they are then used to calculate the required coefficient. A systematic approach, careful algebraic manipulation, and attention to detail, especially with signs, are paramount for solving such problems accurately.