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