What will be the 33rd terms of the sequences 1,7, 25,79,…?

Options

A. `33^3 - 1`
B. `3^(33)-2`
C. `3(33^(3)-2)`
D. none of these

Answer

To find the 33rd term of the sequence 1, 7, 25, 79, we can analyze the pattern in the given terms.

### Step-by-Step Solution:

1. **Identify the Terms**: The first four terms of the sequence are:
- \( a_1 = 1 \)
- \( a_2 = 7 \)
- \( a_3 = 25 \)
- \( a_4 = 79 \)

2. **Look for a Pattern**: We can express these terms in a form that might reveal a pattern. Let's rewrite each term:
- \( a_1 = 3^1 - 2 = 1 \)
- \( a_2 = 3^2 - 2 = 7 \)
- \( a_3 = 3^3 - 2 = 25 \)
- \( a_4 = 3^4 - 2 = 79 \)

3. **Generalize the Pattern**: From the observations above, we can see that the nth term can be expressed as:
\[ a_n = 3^n - 2 \]

4. **Calculate the 33rd Term**: Now, we need to find the 33rd term using the formula derived:
\[ a_{33} = 3^{33} - 2 \]

5. **Final Expression**: Therefore, the 33rd term of the sequence is:
\[ a_{33} = 3^{33} - 2 \]

### Conclusion:
The 33rd term of the sequence is \( 3^{33} - 2 \).