Let's take a look at what happens for every iteration, so we can identify a common pattern that allows us to write down a formula for the n-th iteration.
Let P be the number of people receiving the email.
First iteration (n=1)
Joey sent an email to 2 friends
[tex]P=2[/tex]
Second iteration (n=2)
Those two friends then send it to 3 friends each:
[tex]\begin{gathered} P=2\cdot3 \\ P=2\cdot3^1 \end{gathered}[/tex]
Third iteration (n=3)
Those friends then send it to 3 friends each:
[tex]\begin{gathered} P=2\cdot3\cdot3 \\ P=2\cdot3^2 \end{gathered}[/tex]
Fourth iteration (n=4)
Those friends then send it to 3 friends each:
[tex]\begin{gathered} P=2\cdot3\cdot3\cdot3 \\ P=2\cdot3^3 \end{gathered}[/tex]
Now the pattern becomes clear. For every iteration, we have a 2 multiplying a 3 that has a power equal to the number of the iteration minus one. This way,
n-th iteration:
[tex]P=2\cdot3^{n-1}[/tex]
In terms of the table,
[tex]a_n=2\cdot3^{n-1}[/tex]