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