Key Concept: The Multinomial Theorem

To find the coefficient of a specific term in the expansion of an expression with three or more terms raised to a power, we use the Multinomial Theorem. For an expansion of the form $(x_1 + x_2 + x_3 + \dots + x_k)^n$, the general term is given by:

$\frac{n!}{n_1! n_2! \dots n_k!} x_1^{n_1} x_2^{n_2} \dots x_k^{n_k}$

where $n_1 + n_2 + \dots + n_k = n$.

In this problem, we are expanding $(a + 2b + 4ab)^{10}$. Here, $n=10$, and the terms are $x_1 = a$, $x_2 = 2b$, and $x_3 = 4ab$.

Step-by-Step Derivation of the General Term

1. Identify the terms and their powers: Let the powers of the terms $a$, $2b$, and $4ab$ in the general term be $\alpha$, $\beta$, and $\gamma$ respectively. According to the Multinomial Theorem, the general term (denoted as $T$) in the expansion of $(a + 2b + 4ab)^{10}$ is: $T = \frac{10!}{\alpha! \beta! \gamma!} (a)^\alpha (2b)^\beta (4ab)^\gamma$ Why this step? We use $\alpha, \beta, \gamma$ to represent how many times each term ($a$, $2b$, $4ab$) is chosen from the 10 factors during the expansion. The sum of these powers must equal the total power of the expansion, i.e., $\alpha + \beta + \gamma = 10$.

2. Simplify the general term to collect powers of $a$ and $b$: Now, we distribute the powers and group the terms involving $a$, $b$, and the constants: $T = \frac{10!}{\alpha! \beta! \gamma!} a^\alpha \cdot (2^\beta b^\beta) \cdot (4^\gamma a^\gamma b^\gamma)$ $T = \frac{10!}{\alpha! \beta! \gamma!} (a^{\alpha} a^{\gamma}) (b^{\beta} b^{\gamma}) (2^\beta 4^\gamma)$ $T = \frac{10!}{\alpha! \beta! \gamma!} a^{\alpha+\gamma} b^{\beta+\gamma} 2^\beta (2^2)^\gamma$ $T = \frac{10!}{\alpha! \beta! \gamma!} a^{\alpha+\gamma} b^{\beta+\gamma} 2^\beta 2^{2\gamma}$ $T = \frac{10!}{\alpha! \beta! \gamma!} a^{\alpha+\gamma} b^{\beta+\gamma} 2^{\beta+2\gamma}$ Why this step? We need to identify the combined power of $a$ and $b$ across all terms and isolate the numerical part to match it with the given target term $a^7 b^8$. This simplification helps us create a system of equations.

Setting Up and Solving the System of Equations

We are looking for the coefficient of $a^7 b^8$. By comparing the powers in our simplified general term with the target term, we can form a system of linear equations:

1. Sum of powers: The sum of the individual powers ($\alpha, \beta, \gamma$) must equal the total power of the expansion. $\alpha + \beta + \gamma = 10 \quad \dots(1)$

2. Power of $a$: The combined power of $a$ in the general term must be 7. $\alpha + \gamma = 7 \quad \dots(2)$

3. Power of $b$: The combined power of $b$ in the general term must be 8. $\beta + \gamma = 8 \quad \dots(3)$

Why this step? These equations ensure that we select the correct combination of $\alpha, \beta, \gamma$ that will yield the desired $a^7 b^8$ term.

Now, let's solve the system:

• Solve for $\gamma$: Add equations (2) and (3): $(\alpha + \gamma) + (\beta + \gamma) = 7 + 8$ $\alpha + \beta + 2\gamma = 15 \quad \dots(4)$ Subtract equation (1) from equation (4): $(\alpha + \beta + 2\gamma) - (\alpha + \beta + \gamma) = 15 - 10$ $\gamma = 5$ Why this approach? Adding equations (2) and (3) allows us to introduce $\alpha + \beta + \gamma$, which is known from equation (1), simplifying the system and directly solving for $\gamma$.

• Solve for $\alpha$ and $\beta$: Substitute $\gamma = 5$ into equation (2): $\alpha + 5 = 7 \implies \alpha = 2$

  Substitute $\gamma = 5$ into equation (3): $\beta + 5 = 8 \implies \beta = 3$

  So, the values are $\alpha = 2$, $\beta = 3$, and $\gamma = 5$. Self-check: $\alpha + \beta + \gamma = 2 + 3 + 5 = 10$, which matches equation (1).

Calculating the Coefficient

Now we substitute the values of $\alpha, \beta, \gamma$ into the coefficient part of our general term: The coefficient is $\frac{10!}{\alpha! \beta! \gamma!} 2^{\beta+2\gamma}$. Substitute $\alpha=2, \beta=3, \gamma=5$: $\text{Coefficient} = \frac{10!}{2! 3! 5!} 2^{3+2(5)}$ $\text{Coefficient} = \frac{10!}{2! 3! 5!} 2^{3+10}$ $\text{Coefficient} = \frac{10!}{2! 3! 5!} 2^{13}$

1. Calculate the multinomial coefficient: $\frac{10!}{2! 3! 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{ (2 \times 1) \times (3 \times 2 \times 1) \times 5! }$ $= \frac{10 \times 9 \times 8 \times 7 \times 6}{2 \times 6}$ $= 10 \times 9 \times \frac{8}{2} \times 7 \times \frac{6}{6}$ $= 10 \times 9 \times 4 \times 7 \times 1$ $= 90 \times 28 = 2520$ Why this calculation? This part determines the number of ways to arrange the terms to achieve the desired powers of $a$ and $b$. It's a standard combinatorial calculation.

2. Combine with the power of 2: $\text{Coefficient} = 2520 \times 2^{13}$

Comparing with Given Form and Finding K

The problem states that the coefficient is $K \cdot 2^{16}$. We have calculated the coefficient as $2520 \cdot 2^{13}$. We need to express $2520 \cdot 2^{13}$ in the form $K \cdot 2^{16}$. To do this, we can factor out $2^3$ (which is 8) from 2520: $2520 = 8 \times 315$ Why this step? We need to isolate a $2^{16}$ term from our calculated coefficient to directly find the value of $K$. By expressing $2520$ as a multiple of $2^3$, we can adjust the power of 2.

Now, substitute this back into our coefficient: $\text{Coefficient} = (315 \times 8) \times 2^{13}$ $\text{Coefficient} = 315 \times 2^3 \times 2^{13}$ $\text{Coefficient} = 315 \times 2^{3+13}$ $\text{Coefficient} = 315 \times 2^{16}$

Comparing this with $K \cdot 2^{16}$, we find that: $K = 315$

Important Tips and Common Mistakes

• Careful with exponents: Remember that $(x^m)^n = x^{mn}$. For example, $4^\gamma = (2^2)^\gamma = 2^{2\gamma}$. A common mistake is to write it as $2^{2+\gamma}$ or just $2^\gamma$.
• Correctly identify terms: Ensure you treat each term in the original expansion (e.g., $a$, $2b$, $4ab$) as a distinct unit when applying the multinomial formula, including its constant factor.
• Systematic equation solving: A clear approach to solving the system of equations for $\alpha, \beta, \gamma$ will prevent errors. Always verify that $\alpha + \beta + \gamma = n$.
• Simplification of factorials: Be methodical when canceling terms in factorial expressions to avoid arithmetic errors.

Summary and Key Takeaway

This problem demonstrates the application of the Multinomial Theorem to find coefficients in polynomial expansions. The key steps involve setting up the general term, forming a system of equations based on the desired powers of variables, solving these equations to find the specific powers for each term, calculating the combinatorial and numerical parts of the coefficient, and finally, simplifying the result to match the given form. This systematic approach ensures accuracy in what can be a complex calculation. The answer for K is 315.