The sequence you've provided is:
1, 9, 81, 729, ...
To analyze the sequence, we can look at the relationship between consecutive terms.
- The second term (9) can be expressed as \( 3^2 \).
- The third term (81) can be expressed as \( 9^2 = (3^2)^2 = 3^{2 \times 2} = 3^4 \).
- The fourth term (729) can be expressed as \( 81^2 = (3^4)^2 = 3^{2 \times 4} = 3^8 \).
Now we can see a pattern:
- The first term is \( 3^{0} = 1 \).
- The second term is \( 3^{2} \).
- The third term is \( 3^{4} \).
- The fourth term is \( 3^{8} \).
We can observe that the exponent doubles for each term:
- The exponent starts at \( 0 \) (for the first term), then goes to \( 2 \) (second term), \( 4 \) (third term), and \( 8 \) (fourth term).
- The exponents can be described as \( 0, 2, 4, 8 \), which suggests the exponent follows the pattern \( 2^n \) where \( n = 0, 1, 2, 3, \ldots \).
Thus:
- The first term corresponds to \( 2^0 \) (which is \( 0 \)),
- The second term corresponds to \( 2^1 \) (which is \( 1 \times 2 \)),
- The third term corresponds to \( 2^2 \) (which is \( 2 \times 2 \)),
- The fourth term corresponds to \( 2^3 \) (which is \( 3 \times 2 \)).
In summary, the sequence is a geometric sequence where each term is obtained by squaring the previous term, and the general form can be expressed as:
\[ a_n = 3^{2^{n-1}} \text{ for } n \geq 1. \]
So this sequence can be characterized as a geometric progression where the base is \( 3 \), and the exponent is following a pattern of powers of \( 2 \).