If the coefficient of \( a^7b^8 \) in the expansion of \( (a + 2b + 4ab)^{10} \) is \( K \cdot 2^{16} \), then K is equal to _________.
Show Hint
Always simplify multinomial terms by grouping like powers before setting up your equations. It prevents mistakes when the same variable appears in multiple terms of the base.
Step 1: Understanding the Concept:
The coefficient of a specific term in a multinomial expansion \( (x_1 + x_2 + ... + x_m)^n \) is found using the multinomial theorem. We set up equations for the powers of each variable to solve for the specific indices. Step 2: Key Formula or Approach:
The general term in \( (a + 2b + 4ab)^{10} \) is:
\[ T = \frac{10!}{x!y!z!} (a)^x (2b)^y (4ab)^z = \frac{10!}{x!y!z!} \cdot 2^y \cdot 4^z \cdot a^{x+z} \cdot b^{y+z} \]
where \( x + y + z = 10 \). Step 3: Detailed Explanation:
We are given the term \( a^7b^8 \). Equating the powers:
1. \( x + z = 7 \implies x = 7 - z \)
2. \( y + z = 8 \implies y = 8 - z \)
Substitute into the sum constraint:
\[ (7 - z) + (8 - z) + z = 10 \implies 15 - z = 10 \implies z = 5 \]
Then, \( x = 7 - 5 = 2 \) and \( y = 8 - 5 = 3 \).
The coefficient is:
\[ \text{Coeff} = \frac{10!}{2!3!5!} \cdot 2^3 \cdot (2^2)^5 = \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{2 \cdot 6} \cdot 2^3 \cdot 2^{10} \]
\[ \text{Coeff} = 2520 \cdot 2^{13} \]
We are given this equals \( K \cdot 2^{16} \). So:
\[ 2520 \cdot 2^{13} = K \cdot 2^{16} \implies K = \frac{2520}{2^3} = \frac{2520}{8} = 315 \] Step 4: Final Answer:
The value of K is 315.
