To solve the problem of expanding [tex]\((2m - n)^7\)[/tex], we apply the Binomial Theorem, which states that:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
In our case, [tex]\(a = 2m\)[/tex], [tex]\(b = -n\)[/tex], and [tex]\(n = 7\)[/tex]. Applying the theorem step-by-step:
### Step 1: Initialize the Binomial Expansion
[tex]\[
(2m - n)^7 = \sum_{k=0}^{7} \binom{7}{k} (2m)^{7-k} (-n)^k
\][/tex]
### Step 2: Calculate Each Term
We calculate each term individually:
- For [tex]\(k = 0\)[/tex]:
[tex]\[
\binom{7}{0} (2m)^7 (-n)^0 = 1 \cdot 128m^7 \cdot 1 = 128m^7
\][/tex]
- For [tex]\(k = 1\)[/tex]:
[tex]\[
\binom{7}{1} (2m)^6 (-n)^1 = 7 \cdot 64m^6 \cdot (-n) = -448m^6 n
\][/tex]
- For [tex]\(k = 2\)[/tex]:
[tex]\[
\binom{7}{2} (2m)^5 (-n)^2 = 21 \cdot 32m^5 \cdot n^2 = 672m^5 n^2
\][/tex]
- For [tex]\(k = 3\)[/tex]:
[tex]\[
\binom{7}{3} (2m)^4 (-n)^3 = 35 \cdot 16m^4 \cdot (-n)^3 = -560m^4 n^3
\][/tex]
- For [tex]\(k = 4\)[/tex]:
[tex]\[
\binom{7}{4} (2m)^3 (-n)^4 = 35 \cdot 8m^3 \cdot n^4 = 280m^3 n^4
\][/tex]
- For [tex]\(k = 5\)[/tex]:
[tex]\[
\binom{7}{5} (2m)^2 (-n)^5 = 21 \cdot 4m^2 \cdot (-n)^5 = -84m^2 n^5
\][/tex]
- For [tex]\(k = 6\)[/tex]:
[tex]\[
\binom{7}{6} (2m)^1 (-n)^6 = 7 \cdot 2m \cdot n^6 = 14m n^6
\][/tex]
- For [tex]\(k = 7\)[/tex]:
[tex]\[
\binom{7}{7} (2m)^0 (-n)^7 = 1 \cdot 1 \cdot (-n)^7 = -n^7
\][/tex]
### Step 3: Combine All Terms
Combining all these terms, we get the expansion:
[tex]\[
128m^7 - 448m^6 n + 672m^5 n^2 - 560m^4 n^3 + 280m^3 n^4 - 84m^2 n^5 + 14m n^6 - n^7
\][/tex]
### Conclusion
Therefore, the correct expansion of [tex]\((2m - n)^7\)[/tex] from the provided options is:
[tex]\[
128 m^7 - 448 m^6 n + 672 m^5 n^2 - 560 m^4 n^3 + 280 m^3 n^4 - 84 m^2 n^5 + 14 m n^6 - n^7
\][/tex]
So, the correct choice is:
[tex]\[
\boxed{128 m^7 - 448 m^6 n + 672 m^5 n^2 - 560 m^4 n^3 + 280 m^3 n^4 - 84 m^2 n^5 + 14 m n^6 - n^7}
\][/tex]
