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

2 Answers
Mar 27, 2017

Answer:

#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

Answer:

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.