To find the terms in the binomial expansion of \((a + b)^7\), we can use the binomial theorem, which states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In this case, \( n = 7 \), so the expansion is given by:
\[ (a + b)^7 = \sum_{k=0}^{7} \binom{7}{k} a^{7-k} b^k \]
Each term in the expansion can be written in the form:
\[ \binom{7}{k} a^{7-k} b^k \]
We need to determine whether each of the provided options is a valid term in this expansion:
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Option A: \( a^4b^3 \)
Here, \( 7 - k = 4 \) and \( k = 3 \). Let's check if \( k \) is within bounds:
\[ k = 3 \quad (0 \leq k \leq 7) \Rightarrow \text{Valid} \]
Coefficient:
\[ \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]
Thus, the term is \( 35a^4b^3 \).
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Option B: \( 21a^2b^5 \)
Here, \( 7 - k = 2 \) and \( k = 5 \).
\[ k = 5 \quad (0 \leq k \leq 7) \Rightarrow \text{Valid} \]
Coefficient:
\[ \binom{7}{5} = \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21 \]
Thus, the term is \( 21a^2b^5 \).
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Option C: \( 21a^2b^4 \)
Here, \( 7 - k = 2 \) and \( k = 4 \) leads to:
\[ k = 4 \quad (0 \leq k \leq 7) \Rightarrow \text{Valid} \]
Coefficient:
\[ \binom{7}{4} = \binom{7}{3} = 35 \]
Thus, the term is \( 35a^2b^4 \).
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Option D: \( 21a^6b \)
Here, \( 7 - k = 6 \) and \( k = 1 \).
\[ k = 1 \quad (0 \leq k \leq 7) \Rightarrow \text{Valid} \]
Coefficient:
\[ \binom{7}{1} = 7 \]
Thus, the term is \( 7a^6b \).
Based on this analysis, the terms that are valid in the expansion of \((a + b)^7\) are:
- \(A: 35a^4b^3\) (valid, but coefficient is 35)
- \(B: 21a^2b^5\) (valid)
- \(C: 35a^2b^4\) (valid, but coefficient is 35)
- \(D: 7a^6b\) (valid, but coefficient is 7)
Therefore, the correct answer with the stated coefficient is:
B. \(21a^2b^5\).
1 answer