Question

Pritika is using the binomial theorum to find the binomial expansion of (a+b)^7. Which of the following is a term in the expansion? 21a^6b, a^4b^3, 21a^2b^5, or 21a^2b^4?

Answer 1

To find out which term in the expansion of \((a + b)^7\) is valid, we can use the Binomial Theorem, which states that:

\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]

For \((a + b)^7\), we have \(n = 7\). Thus, the general term in the expansion can be represented as:

\[
T_k = \binom{7}{k} a^{7-k} b^k
\]

where \(k = 0, 1, 2, ... , 7\).

To determine if each term provided in the options is valid, we will check the exponents of \(a\) and \(b\) in the terms, as well as confirm that they can be expressed in the form \( \binom{7}{k} a^{7-k} b^k \).

1. **Checking \(21a^6b\)**:
- This corresponds to \(a^{7-k} = a^6\) (so, \(7-k = 6\) means \(k=1\)) and \(b^k = b^1\).
- We have \(k = 1\), thus the coefficient should be \(\binom{7}{1} = 7\), leading to the term \(7a^6b\).
- **Not valid** (the coefficient is not 21).

2. **Checking \(a^4b^3\)**:
- This corresponds to \(a^{7-k} = a^4\) (hence \(7-k = 4\) means \(k=3\)) and \(b^k = b^3\).
- We have \(k = 3\), hence the coefficient should be \(\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\), resulting in \(35a^4b^3\).
- **Not valid** (the coefficient is not 1).

3. **Checking \(21a^2b^5\)**:
- This corresponds to \(a^{7-k} = a^2\) (so \(7-k = 2\) means \(k=5\)) and \(b^k = b^5\).
- We have \(k = 5\), thus the coefficient should be \(\binom{7}{5} = \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21\), resulting in \(21a^2b^5\).
- **Valid term**.

4. **Checking \(21a^2b^4\)**:
- This corresponds to \(a^{7-k} = a^2\) (so \(7-k = 2\) means \(k=5\)) and \(b^k = b^4\).
- Here, \(k\) would need to be 4, which does not fit the previous logic for \(a^2\).
- **Not valid** (the powers do not match).

The only correct term from the options given that appears in the binomial expansion of \((a+b)^7\) is:

**21a^2b^5**.