How do you find the nth term of the sequence 2, 4, 16, 256, ...?

2 Answers
Mar 27, 2017

a_n = 2^(2^(n))

Explanation:

Okay, we have 2, 4, 16, 256, ....?
Find a common difference between each one. They are all divisible by 2. Since they are all divisible by 2, we have 2^1, 2^2, 2^4, 2^8, ...?
Now let's look at the power exponents, 1, 2, 4, 8,...?
It looks like for 1, 2, 4, 8, ...? can work if we have 2^n, starting a 0.
Now we have 2^(2^(n))

Plug in to be sure:
2^(2^(0)) = 2
2^(2^(1)) = 4
2^(2^(3)) = 16
2^(2^(4)) = 256

Mar 29, 2017

It could be a_n = 2^(2^n) or any matching formula.

Explanation:

Given:

2, 4, 16, 256,...

It seems that each element of the sequence is the square of the preceding one, since:

4 = 2^2

16 = 4^2

256 = 16^2

This would result in the formula:

a_n = 2^(2^n)

However, note that we have been told nothing about the nature of this sequence except the first 4 terms. We have not even been told that it is a sequence of numbers.

For example, it can be matched with a cubic formula:

a_n = 1/3 (109n^3 - 639n^2 + 1160n - 624)

Then it would not follow the squaring pattern, but would continue:

2, 4, 16, 256, 942, 2292, 4524,...

We could choose any following numbers we like and find a formula that matches them.

No infinite sequence is determined purely by its first few terms.