Question

Pritika is using the Binomial Theorem to find the binomial expansion of (a+b)7 . Which of the following is a term in the expansion.(1 point) Responses 21a6b 21 A to the 6th power b 21a2b4 21 A squared b to the 4th power a4b3 A to the 4th power b cubed 21a2b5

Answer 1

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:

1. **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 \(21a^6b\).

2. **For \(21 a^6 b\)**: The same reasoning applies as above. It is \(7 a^6 b\) which does not equal \(21 a^6 b\).

3. **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 **\(21a^2b^4\)**.

4. **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 \(21 a^4 b^3\).

5. **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.**