To find the coefficient of the [tex]\( x^4 y^3 \)[/tex] term in the expansion of [tex]\((x - 2y)^7\)[/tex], we can use the Binomial Theorem. The Binomial Theorem states:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
Here, our expression is [tex]\((x - 2y)^7\)[/tex]. We need to find the specific term that contains [tex]\( x^4 y^3 \)[/tex]. First, rewrite the given expression in the form of [tex]\((a + b)^n\)[/tex]:
[tex]\[
(x - 2y)^7
\][/tex]
We can identify [tex]\( a = x \)[/tex], [tex]\( b = -2y \)[/tex], and [tex]\( n = 7 \)[/tex].
Using the Binomial Theorem, the general term in the expansion is:
[tex]\[
\binom{7}{k} x^{7-k} (-2y)^k
\][/tex]
We are looking for the term where the powers of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] match [tex]\( x^4 \)[/tex] and [tex]\( y^3 \)[/tex] respectively. This gives us two conditions for the exponents:
1. [tex]\( 7 - k = 4 \)[/tex] (for [tex]\( x^4 \)[/tex])
2. [tex]\( k = 3 \)[/tex] (for [tex]\( y^3 \)[/tex])
By solving [tex]\( 7 - k = 4 \)[/tex], we get:
[tex]\[
k = 3
\][/tex]
Now, substitute [tex]\( k = 3 \)[/tex] into the general term formula:
[tex]\[
\binom{7}{3} x^{7-3} (-2y)^3
\][/tex]
Simplify the expression:
[tex]\[
\binom{7}{3} x^4 (-2y)^3
\][/tex]
Calculate the binomial coefficient:
[tex]\[
\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35
\][/tex]
Next, calculate [tex]\((-2y)^3\)[/tex]:
[tex]\[
(-2y)^3 = (-2)^3 \cdot y^3 = -8y^3
\][/tex]
Combine these results into the term we found:
[tex]\[
35 \cdot x^4 \cdot (-8y^3) = 35 \cdot (-8) \cdot x^4 y^3
\][/tex]
Simplify:
[tex]\[
35 \cdot (-8) = -280
\][/tex]
Therefore, the coefficient of the [tex]\( x^4 y^3 \)[/tex] term in the expansion of [tex]\((x - 2y)^7\)[/tex] is:
[tex]\[
\boxed{-280}
\][/tex]
