nyungu6f5

2023-03-10

How to find the nth term of the sequence $2,4,16,256,...$?

enkestaxn2

Okay, we have $2,4,16,256,....?$
Look for a shared distinction between each. 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:
${2}^{{2}^{0}}=2$
${2}^{{2}^{1}}=4$
${2}^{{2}^{3}}=16$
${2}^{{2}^{4}}=256$

posaminehri

Given:

$2,4,16,256,...$

Each element of the sequence appears to be the square of the one before it, as shown by:

$4={2}^{2}$
$16={4}^{2}$
$256={16}^{2}$

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:

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

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.

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