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

##### 2 Answers

#### Answer:

#### Explanation:

Okay, we have

Find a common difference between each one. They are all divisible by

Now let's look at the power exponents,

It looks like for

Now we have

Plug in to be sure:

#### Answer:

It could be

#### 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

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.**