Which of the following is the expansion of $(2 m-n)^7$ ?

A$128 m^7+448 m^6 n+672 m^5 n^2+560 m^4 n^3+280 m^3 n^4+84 m^2 n^5+14 m n^6+n^7$
B$128 m^7+14 m^6 n+42 m^5 n^2+70 m^4 n^3+70 m^3 n^4+42 m^2 n^5+14 m n^6+n^7$
C$m^7+7 m^6 n+21 m^5 n^2+35 m^4 n^3+35 m^3 n^4+21 m^2 n^5+7 m n^6+n^7$
D$128 m^7-448 m^6 n+672 m^5 n^2-560 m^4 n^3+280 m^3 n^4-84 m^2 n^5+14 m n^6-n^7$
E$128 m^7-384 m^6 n+480 m^5 n^2-320 m^4 n^3+160 m^3 n^4-60 m^2 n^5+12 m n^6-n^7$

Question

05-11-2026
Mathematics

Answered

Which of the following is the expansion of $(2 m-n)^7$ ?

A$128 m^7+448 m^6 n+672 m^5 n^2+560 m^4 n^3+280 m^3 n^4+84 m^2 n^5+14 m n^6+n^7$
B$128 m^7+14 m^6 n+42 m^5 n^2+70 m^4 n^3+70 m^3 n^4+42 m^2 n^5+14 m n^6+n^7$
C$m^7+7 m^6 n+21 m^5 n^2+35 m^4 n^3+35 m^3 n^4+21 m^2 n^5+7 m n^6+n^7$
D$128 m^7-448 m^6 n+672 m^5 n^2-560 m^4 n^3+280 m^3 n^4-84 m^2 n^5+14 m n^6-n^7$
E$128 m^7-384 m^6 n+480 m^5 n^2-320 m^4 n^3+160 m^3 n^4-60 m^2 n^5+12 m n^6-n^7$

Verified Answer :

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GinnyAnswer GinnyAnswer · Answer 1 · 2026-11-05T14:23:12-05:00

The binomial theorem is used to expand $(2m - n)^7$ where $a=2m$, $b=-n$, and $n=7$.
Terms are computed by calculating $\binom{7}{k} (2m)^{7-k} (-n)^k$ for $k=0, 1, \dots, 7$.
Signs alternate because the second term is negative, resulting in the sequence: $128m^7 - 448m^6n + 672m^5n^2 - 560m^4n^3 + 280m^3n^4 - 84m^2n^5 + 14mn^6 - n^7$.
This matches option d.
Final answer: $\boxed{d}$

Explanation

Applying the Binomial Theorem

The expansion of a binomial power $(a + b)^n$ is given by the Binomial Theorem:
\[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k,\]
where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient. For the expression $(2m - n)^7$, we identify $a = 2m$, $b = -n$, and $n = 7$. Substituting these values into the theorem gives us:
\[(2m - n)^7 = \sum_{k=0}^{7} \binom{7}{k} (2m)^{7-k} (-n)^k.\]
The coefficients for $n=7$ are determined by the 7th row of Pascal's triangle: $1, 7, 21, 35, 35, 21, 7, 1$. Because $b = -n$, the terms in the expansion will alternate in sign, starting with a positive sign.

Calculating Individual Terms

We calculate each term of the expansion for $k = 0$ to $7$:

For $k=0$: $\binom{7}{0} (2m)^7 (-n)^0 = 1 \cdot 128m^7 \cdot 1 = 128m^7$
For $k=1$: $\binom{7}{1} (2m)^6 (-n)^1 = 7 \cdot 64m^6 \cdot (-n) = -448m^6n$
For $k=2$: $\binom{7}{2} (2m)^5 (-n)^2 = 21 \cdot 32m^5 \cdot n^2 = 672m^5n^2$
For $k=3$: $\binom{7}{3} (2m)^4 (-n)^3 = 35 \cdot 16m^4 \cdot (-n^3) = -560m^4n^3$
For $k=4$: $\binom{7}{4} (2m)^3 (-n)^4 = 35 \cdot 8m^3 \cdot n^4 = 280m^3n^4$
For $k=5$: $\binom{7}{5} (2m)^2 (-n)^5 = 21 \cdot 4m^2 \cdot (-n^5) = -84m^2n^5$
For $k=6$: $\binom{7}{6} (2m)^1 (-n)^6 = 7 \cdot 2m \cdot n^6 = 14mn^6$
For $k=7$: $\binom{7}{7} (2m)^0 (-n)^7 = 1 \cdot 1 \cdot (-n^7) = -n^7$

Final Result and Comparison

Combining all the terms, the full expansion is:
\[128m^7 - 448m^6n + 672m^5n^2 - 560m^4n^3 + 280m^3n^4 - 84m^2n^5 + 14mn^6 - n^7.\]
Comparing this result with the provided options, we see that it matches option d. Therefore, the expansion of $(2m - n)^7$ is $\boxed{d}$.

Examples

Binomial expansions like this are essential in fields such as probability, statistics, and computer science. For example, if you are modeling the probability of success in a sequence of events where you have a "2-to-1" ratio of outcomes (like weightings in a game or a resource allocation problem), expanding the power of the sum of these variables allows you to calculate the distribution of all possible combined outcomes. By using the binomial theorem, you can efficiently find the probability of exactly $k$ successes out of $n$ trials without needing to manually compute every single outcome tree!