The question is asking for the coefficient of $x^9$ in the expansion of $(2 - x)^{19}$.

To find the coefficient of $x^9$ in this expansion, we can use the Binomial Theorem, which states:

$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

In this case:

• $a = 2$
• $b = -x$
• $n = 19$

We are looking for the term containing $x^9$, so we need the term where $k = 9$.

The general term in the expansion is: $\binom{19}{k} \cdot (2)^{19-k} \cdot (-x)^k$

Substituting $k = 9$: $\binom{19}{9} \cdot (2)^{19-9} \cdot (-x)^9$

Simplifying this expression:

1. $\binom{19}{9}$ gives the binomial coefficient for $k = 9$.
2. $(2)^{10}$ represents the power of 2.
3. $(-x)^9$ includes $(-1)^9 = -1$ and $x^9$.

So, the coefficient of $x^9$ is: $-\binom{19}{9} \cdot 2^{10}$

Calculating:

1. $\binom{19}{9} = 92378$
2. $2^{10} = 1024$

Thus, the coefficient of $x^9$ is: $-92378 \cdot 1024 = -94633984$

Final Answer:

The coefficient of $x^9$ in the expansion of $(2 - x)^{19}$ is $-94633984$.

Would you like further details, or do you have any questions?

Here are five related questions to expand your understanding:

1. How would the coefficient change if the power were different, say $(2 - x)^{20}$?
2. What is the significance of the negative sign in this coefficient?
3. How does the Binomial Theorem apply to more complex expressions, such as $(3x + y)^{10}$?
4. Can we generalize this method for any exponent and any specific power of $x$?
5. How would the answer differ if we were finding the coefficient of $x^{10}$ instead?

Tip: Remember that the sign of each term in a binomial expansion alternates based on the powers of negative terms in the expression.