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

Question

Christopher Allers · Accepted Answer

#a_n = 2^(2^(n))# 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#

Luke Alvarez · Answer

It could be #a_n = 2^(2^n)# or any matching formula.

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.

Luke Alvarez · Answer

undefined To find the nth term of the sequence 2, 4, 16, 256, ..., you can observe that each term is obtained by raising 2 to the power of the previous term. Therefore, the nth term can be expressed as $2^{2^{n-1}}$.