Writeanequationtodescribethesequencebelow.Usentorepresentthepositionofaterminthesequence,wheren = 1forthefirstterm.

3 , 6 , 12 , ...
Write your answer using decimals and integers.

an = _(_)^ n^ -^ 1

1 answer

To derive an equation for the sequence 3, 6, 12, ..., let's first examine how the terms progress.

  1. Identifying the pattern:
    • The first term \( a_1 = 3 \)
    • The second term \( a_2 = 6 \)
    • The third term \( a_3 = 12 \)

It appears that each term can be obtained from the previous term by multiplying. Specifically:

  • \( 6 = 3 \times 2 \)
  • \( 12 = 6 \times 2 \)

From this, we can see that each term is double the previous term. This suggests a relationship involving powers of 2.

  1. General formula: The first term can be expressed as \( 3 = 3 \times 2^0 \).
    • The second term can be expressed as \( 6 = 3 \times 2^1 \).
    • The third term can be expressed as \( 12 = 3 \times 2^2 \).

Thus, the pattern here is \( a_n = 3 \times 2^{n-1} \), where \( n \) is the position in the sequence (with \( n = 1 \) for the first term).

  1. Final equation: So, the equation that describes the sequence is:

\[ a_n = 3 \times 2^{n-1} \]

This matches the given structure \( a_n = ()^n - 1 \) if we manipulate it slightly, but here we maintain the relationship as discovered.

Thus, the final equation for the sequence is:

\[ a_n = 3 \times 2^{n-1} \]