Find the first 4 terms and the 20th term of the sequence whose
`S_(n) = (3)/(2) (3^(n)-1).`

Answer

To find the first four terms and the 20th term of the sequence defined by the sum of the first n terms \( S_n = \frac{3}{2}(3^n - 1) \), we can use the relationship between the sum of terms and the individual terms of the sequence.

### Step-by-Step Solution:

1. **Understanding the Formula**:
The nth term \( a_n \) of the sequence can be found using the formula:
\[ a_n = S_n - S_{n-1} \]
where \( S_n \) is the sum of the first n terms.

2. **Finding the First Term \( a_1 \)**:
To find \( a_1 \):
\[ a_1 = S_1 = S_1 - S_0 \]
Since \( S_0 = 0 \) (the sum of zero terms), we have:
\[ S_1 = \frac{3}{2}(3^1 - 1) = \frac{3}{2}(3 - 1) = \frac{3}{2} \times 2 = 3 \]
Thus, \( a_1 = 3 \).

3. **Finding the Second Term \( a_2 \)**:
To find \( a_2 \):
\[ a_2 = S_2 - S_1 \]
Calculate \( S_2 \):
\[ S_2 = \frac{3}{2}(3^2 - 1) = \frac{3}{2}(9 - 1) = \frac{3}{2} \times 8 = 12 \]
Now, substituting back:
\[ a_2 = S_2 - S_1 = 12 - 3 = 9 \]

4. **Finding the Third Term \( a_3 \)**:
To find \( a_3 \):
\[ a_3 = S_3 - S_2 \]
Calculate \( S_3 \):
\[ S_3 = \frac{3}{2}(3^3 - 1) = \frac{3}{2}(27 - 1) = \frac{3}{2} \times 26 = 39 \]
Now, substituting back:
\[ a_3 = S_3 - S_2 = 39 - 12 = 27 \]

5. **Finding the Fourth Term \( a_4 \)**:
To find \( a_4 \):
\[ a_4 = S_4 - S_3 \]
Calculate \( S_4 \):
\[ S_4 = \frac{3}{2}(3^4 - 1) = \frac{3}{2}(81 - 1) = \frac{3}{2} \times 80 = 120 \]
Now, substituting back:
\[ a_4 = S_4 - S_3 = 120 - 39 = 81 \]

6. **Finding the 20th Term \( a_{20} \)**:
To find \( a_{20} \):
\[ a_{20} = S_{20} - S_{19} \]
Calculate \( S_{20} \):
\[ S_{20} = \frac{3}{2}(3^{20} - 1) \]
Calculate \( S_{19} \):
\[ S_{19} = \frac{3}{2}(3^{19} - 1) \]
Now substituting:
\[ a_{20} = S_{20} - S_{19} = \left(\frac{3}{2}(3^{20} - 1)\right) - \left(\frac{3}{2}(3^{19} - 1)\right) \]
Simplifying:
\[ a_{20} = \frac{3}{2}(3^{20} - 1 - 3^{19} + 1) = \frac{3}{2}(3^{20} - 3^{19}) = \frac{3}{2}(3^{19}(3 - 1)) = \frac{3}{2}(2 \cdot 3^{19}) = 3^{20} \]

### Summary of Results:
- The first four terms are:
- \( a_1 = 3 \)
- \( a_2 = 9 \)
- \( a_3 = 27 \)
- \( a_4 = 81 \)
- The 20th term is:
- \( a_{20} = 3^{20} \)