Question

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

Answer 1

`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`

Answer 2

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.