In the expansion of `(a+b)^(n)`, first three terms are 243, 810 and 1080 respectively, then the fourth term of the expansion is `(n in N)`

Options

A. 32
B. 720
C. 510
D. 420

Answer

To solve the problem, we need to find the fourth term in the expansion of \((a + b)^n\) given that the first three terms are 243, 810, and 1080 respectively.

### Step-by-Step Solution:

1. **Identify the first three terms**:
The first three terms of the binomial expansion of \((a + b)^n\) are:
- First term: \(T_1 = \binom{n}{0} a^n b^0 = a^n\)
- Second term: \(T_2 = \binom{n}{1} a^{n-1} b^1 = n a^{n-1} b\)
- Third term: \(T_3 = \binom{n}{2} a^{n-2} b^2 = \frac{n(n-1)}{2} a^{n-2} b^2\)

Given:
- \(T_1 = 243\)
- \(T_2 = 810\)
- \(T_3 = 1080\)

2. **Set up equations**:
From the first term:
\[ a^n = 243 \quad \text{(1)} \]
From the second term:
\[ n a^{n-1} b = 810 \quad \text{(2)} \]
From the third term:
\[ \frac{n(n-1)}{2} a^{n-2} b^2 = 1080 \quad \text{(3)} \]

3. **Solve for \(a\)**:
From equation (1), we can express \(a\) in terms of \(n\):
\[ a = 243^{1/n} \]

4. **Substitute \(a\) into equation (2)**:
Substitute \(a\) into equation (2):
\[ n (243^{1/n})^{n-1} b = 810 \]
Simplifying gives:
\[ n \cdot 243^{(n-1)/n} b = 810 \]
This can be rearranged to find \(b\):
\[ b = \frac{810}{n \cdot 243^{(n-1)/n}} \quad \text{(4)} \]

5. **Substitute \(a\) and \(b\) into equation (3)**:
Substitute \(a\) and \(b\) into equation (3):
\[ \frac{n(n-1)}{2} (243^{1/n})^{n-2} \left(\frac{810}{n \cdot 243^{(n-1)/n}}\right)^2 = 1080 \]
Simplifying this equation will yield a relationship between \(n\) and constants.

6. **Finding \(n\)**:
After simplifying the above equation, we can solve for \(n\). Through calculations, we find that \(n = 5\).

7. **Finding \(a\) and \(b\)**:
Substitute \(n = 5\) back into equation (1):
\[ a^5 = 243 \implies a = 3 \]
Substitute \(n = 5\) into equation (4):
\[ b = \frac{810}{5 \cdot 243^{4/5}} = 2 \]

8. **Calculate the fourth term**:
The fourth term \(T_4\) is given by:
\[ T_4 = \binom{n}{3} a^{n-3} b^3 \]
Substituting \(n = 5\), \(a = 3\), and \(b = 2\):
\[ T_4 = \binom{5}{3} (3)^{2} (2)^{3} \]
Calculating this gives:
\[ T_4 = 10 \cdot 9 \cdot 8 = 720 \]

### Final Answer:
The fourth term of the expansion is \(720\).