Consider the sequence 2, 6, 18, 54,...
Let n = the term number in the sequence.
Let A(n) = the value of the nth term in the sequence.
A(1) = 2 = 2 X 3^0
A(2) = 6 = 2 X 3 = 2 X 3^1
A(3) = 18 = 2 X 3 X 3 = 2 X 3^2
A(4) = 54 = 2 X 3 X 3 X 3 = 2 X 3^3
What is the relationship between the exponent of the base 3 and the value of n?
Well, it seems like the exponent of the base 3 is actually equal to n minus 1. So if we have A(n), the value of the nth term in the sequence, the exponent of the base 3 will be n minus 1. It's like the base 3 just can't keep up with n and always lags behind by 1. I guess it's just not as fast as we thought!
The relationship between the exponent of the base 3 and the value of n is that the exponent is equal to n minus one.
In this sequence, notice that each term is obtained by multiplying the previous term by 3. The first term, A(1), can be expressed as 2 X 3^0, where the exponent 0 is equal to (1 - 1).
Similarly, the second term, A(2), is obtained by multiplying the first term by 3. So we can write A(2) = 2 X 3^1, where the exponent 1 is equal to (2 - 1).
The pattern continues for the third and fourth terms. A(3) = 2 X 3^2, with the exponent 2 equal to (3 - 1). A(4) = 2 X 3^3, with the exponent 3 equal to (4 - 1).
Therefore, for any term in this sequence, the exponent of the base 3 is equal to n minus one.