Key Concepts and Formulas

This problem relies on the Binomial Theorem, specifically its application for finding the initial terms of a binomial expansion. For any positive integer $n$, the binomial expansion of $(1+x)^n$ and $(1-x)^n$ up to the term $x^2$ are given by:

1. For $(1+x)^n$: $(1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots$ Which simplifies to: $(1+x)^n = 1 + nx + \frac{n(n-1)}{2}x^2 + O(x^3)$ Here, $O(x^3)$ represents terms with $x^3$ and higher powers, which are not relevant for finding coefficients of $x$ and $x^2$.

2. For $(1-x)^n$: We can obtain this by substituting $(-x)$ for $x$ in the expansion of $(1+x)^n$: $(1-x)^n = 1 + n(-x) + \frac{n(n-1)}{2}(-x)^2 + O(x^3)$ Which simplifies to: $(1-x)^n = 1 - nx + \frac{n(n-1)}{2}x^2 + O(x^3)$ Notice the alternating signs, which is a crucial detail for $(1-x)^n$.

Problem Statement

We are given an expression $(1+x)^p(1-x)^q$. We know that the coefficient of $x$ in its expansion is $4$ and the coefficient of $x^2$ is $-5$. Our objective is to determine the value of the expression $2p+3q$.

Step-by-Step Solution

1. Expand Each Binomial Factor

Our first step is to expand each part of the product, $(1+x)^p$ and $(1-x)^q$, using the binomial theorem, but only up to the $x^2$ term. We do this because any terms with $x^3$ or higher powers will not contribute to the coefficients of $x$ or $x^2$ in the final product.

Applying the formulas above:

• For $(1+x)^p$: $(1+x)^p = 1 + px + \frac{p(p-1)}{2}x^2 + \dots$
• For $(1-x)^q$: $(1-x)^q = 1 - qx + \frac{q(q-1)}{2}x^2 + \dots$

2. Multiply the Expanded Series to Find the Product's Expansion

Now, we multiply these two truncated series. We need to be careful to only collect terms that contribute to the coefficients of $x$ and $x^2$.

$(1+x)^p(1-x)^q = \left(1 + px + \frac{p(p-1)}{2}x^2 + \dots \right) \left(1 - qx + \frac{q(q-1)}{2}x^2 + \dots \right)$

Let's systematically identify the terms for each power of $x$:

• Constant term (coefficient of $x^0$): This comes from multiplying the constant terms of each expansion: $1 \times 1 = 1$.

• Terms contributing to the coefficient of $x$ ($x^1$): These arise from multiplying a constant term by an $x$ term, or an $x$ term by a constant term.

  • From $(1+x)^p$: $px$ (multiplied by $1$ from $(1-x)^q$)
  • From $(1-x)^q$: $-qx$ (multiplied by $1$ from $(1+x)^p$) Combining these: $px - qx = (p-q)x$. Therefore, the coefficient of $x$ is $(p-q)$.

• Terms contributing to the coefficient of $x^2$ ($x^2$): These terms can arise from three types of multiplications:

  1. Constant term from first expansion $\times$ $x^2$ term from second expansion: $1 \times \left(\frac{q(q-1)}{2}x^2\right) = \frac{q(q-1)}{2}x^2$
  2. $x$ term from first expansion $\times$ $x$ term from second expansion: $(px) \times (-qx) = -pqx^2$
  3. $x^2$ term from first expansion $\times$ Constant term from second expansion: $\left(\frac{p(p-1)}{2}x^2\right) \times 1 = \frac{p(p-1)}{2}x^2$

  Combining these $x^2$ terms: $\left(\frac{q(q-1)}{2} - pq + \frac{p(p-1)}{2}\right)x^2$ Therefore, the coefficient of $x^2$ is $\left(\frac{p(p-1)}{2} - pq + \frac{q(q-1)}{2}\right)$.

3. Formulate Equations from Given Information

The problem provides specific values for these coefficients. We will use them to set up a system of equations.

• Equation for coefficient of $x$: Given that the coefficient of $x$ is $4$: $p - q = 4 \quad \text{.........(i)}$

• Equation for coefficient of $x^2$: Given that the coefficient of $x^2$ is $-5$: $\frac{p(p-1)}{2} - pq + \frac{q(q-1)}{2} = -5$ To simplify this equation, multiply the entire equation by $2$ to eliminate the denominators: $p(p-1) - 2pq + q(q-1) = -10$ Expand the terms: $p^2 - p - 2pq + q^2 - q = -10$ Rearrange the terms to group common patterns, specifically noticing the perfect square $(p-q)^2$: $(p^2 - 2pq + q^2) - (p+q) = -10$ Substitute $(p-q)^2$ for $(p^2 - 2pq + q^2)$: $(p-q)^2 - (p+q) = -10 \quad \text{.........(ii)}$ This rearrangement simplifies the equation significantly and allows us to use equation (i).

4. Solve the System of Equations for $p$ and $q$

We now have two critical equations:

1. $p - q = 4$
2. $(p-q)^2 - (p+q) = -10$

Substitute the value of $(p-q)$ from equation (i) into equation (ii): $(4)^2 - (p+q) = -10$ $16 - (p+q) = -10$ Now, isolate $(p+q)$: $-(p+q) = -10 - 16$ $-(p+q) = -26$ $p+q = 26 \quad \text{.........(iii)}$

We now have a simpler system of two linear equations:

1. $p - q = 4$
2. $p + q = 26$

To solve for $p$ and $q$:

• Add Equation (i) and Equation (iii): This eliminates $q$: $(p - q) + (p + q) = 4 + 26$ $2p = 30$ $p = 15$

• Subtract Equation (i) from Equation (iii): This eliminates $p$: $(p + q) - (p - q) = 26 - 4$ $p + q - p + q = 22$ $2q = 22$ $q = 11$

So, we have found that $p=15$ and $q=11$.

5. Calculate the Final Expression $2p+3q$

The problem asks for the value of $2p+3q$. We substitute the values of $p=15$ and $q=11$ that we just found: $2p + 3q = 2(15) + 3(11)$ $= 30 + 33$ $= 63$

Tips and Common Mistakes to Avoid

• Sign Errors in $(1-x)^q$: A frequent error is forgetting the alternating signs when expanding $(1-x)^q$. Always remember it's $1 - qx + \frac{q(q-1)}{2}x^2$, not $1 + qx + \dots$.
• Careless Algebraic Manipulation: The step where you simplify the coefficient of $x^2$ is prone to errors. Be meticulous when expanding $p(p-1)$, $q(q-1)$, and collecting like terms. Double-check your signs and arithmetic.
• Missing Terms in Multiplication: When multiplying the two series, ensure all combinations that yield $x$ or $x^2$ terms are included. A systematic approach (e.g., term by term) helps avoid omissions.
• Recognizing Algebraic Identities: The simplification from $p^2 - p - 2pq + q^2 - q$ to $(p-q)^2 - (p+q)$ is a crucial step that makes the problem much easier. Always look for perfect squares or other identities to simplify expressions.

Summary and Key Takeaway

This problem demonstrates a classic application of the Binomial Theorem in finding specific coefficients of a polynomial product. The core strategy involves:

1. Performing partial binomial expansions for each factor up to the necessary power.
2. Multiplying these partial expansions systematically.
3. Equating the resulting coefficients to the given values to form a system of equations.
4. Carefully solving the system of equations. Success hinges on precision in binomial expansion, accurate algebraic manipulation, and recognizing opportunities to simplify equations.