To find the binomial expansion of \((a + b)^7\), we can use the Binomial Theorem, which states:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In this case, \(n = 7\). The general term in the expansion is given by:
\[ T_k = \binom{7}{k} a^{7-k} b^k \]
Now, we can determine which of the provided terms could be in the expansion:
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For \(21a^6b\): Here, \(a^{7-k} = a^6\) means \(7-k = 6\) or \(k = 1\). We then have: \[ T_1 = \binom{7}{1} a^6 b^1 = 7 a^6 b. \] This term does not match \(21a^6b\).
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For \(21 a^6 b\): The same reasoning applies as above. It is \(7 a^6 b\) which does not equal \(21 a^6 b\).
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For \(21a^2b^4\): Here, \(a^{7-k} = a^2\) means \(7-k = 2\) or \(k = 5\). We then calculate: \[ T_5 = \binom{7}{5} a^2 b^5 = \binom{7}{2} a^2 b^5 = 21 a^2 b^5, \] Incorrect notation for \(21a^2b^4\).
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For \(21 a^4 b^3\): Here, \(a^{7-k} = a^4\) means \(7-k = 4\) or \(k = 3\). We calculate: \[ T_3 = \binom{7}{3} a^4 b^3 = 35 a^4 b^3. \] This does not equal \(21 a^4 b^3\).
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For \(21 a^2 b^5\): As calculated above, \[ T_5 = 21 a^2 b^5. \] This matches.
Therefore, the correct term in the expansion is:
21a^2b^5.
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