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

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:

$2 , 4 , 16 , 256 , \ldots$

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:

${a}_{n} = \frac{1}{3} \left(109 {n}^{3} - 639 {n}^{2} + 1160 n - 624\right)$

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

$2 , 4 , 16 , 256 , 942 , 2292 , 4524 , \ldots$

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.