Question

pritika is using the binomial theorem to find tbe binomial expansion of (a+b)^7 which of the following is a term in the expansion
A. A^4b^3
B. 21a^2b^5
C. 21a^2b^4
D. 21a^6b

Answer 1

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:

1. **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 \( 35a^4b^3 \).

2. **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 \( 21a^2b^5 \).

3. **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 \( 35a^2b^4 \).

4. **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. \(21a^2b^5\)**.