Question

what is the coefficient of the x^4y^3 term in the expansion of (x - 2y)^7

Answer 1

To find the coefficient of the \(x^4y^3\) term in the expansion of \((x - 2y)^7\), we can use the Binomial Theorem. The theorem states that:

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

In our case, \(a = x\), \(b = -2y\), and \(n = 7\). We want the term that corresponds to \(x^4y^3\).

In the expansion, the general term can be written as:

\[
T_k = \binom{n}{k} a^{n-k} b^k
\]

Substituting \(n = 7\), \(a = x\), and \(b = -2y\), we have:

\[
T_k = \binom{7}{k} x^{7-k} (-2y)^k
\]

We need to find \(k\) such that the term contains \(x^4\) and \(y^3\).

From the term \(x^{7-k}\):
\[
7 - k = 4 \implies k = 3
\]

Now substitute \(k = 3\) into the general term to find the coefficient:

\[
T_3 = \binom{7}{3} x^{7-3} (-2y)^3
\]
\[
= \binom{7}{3} x^4 (-2)^3 y^3
\]

Calculating \(\binom{7}{3}\):

\[
\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = \frac{210}{6} = 35
\]

Now calculate \((-2)^3\):

\[
(-2)^3 = -8
\]

Now substituting back we find:

\[
T_3 = 35 x^4 (-8) y^3 = -280 x^4 y^3
\]

Therefore, the coefficient of the \(x^4y^3\) term is:

\[
\boxed{-280}
\]