I am trying to conjecture a formula for the $n$th term of $\{a_n\}$ if the first ten terms of the sequence are as follows
1.) $$2, 6, 18, 54, 162, 486, 1458, 4374, 13122, 39366$$
2.) $$1, 1, 0, 1, 1, 0, 1, 1, 0, 1$$
I have no idea how to even start to come up with a formula. Do I need to use induction here?
For the first, we have $a_0 = 2,$ and for $n\in \mathbb Z, n\geq 1,\quad a_n = 3a_{n-1}$
Alternatively, note that $\dfrac{a_n}{a_{n-1}} = 3$.
For the second, if we start with $n=1$, so $a_1 = 1$, then for $n\geq 1$ define $$a_n = \begin{cases} 1 & n\not\equiv 0 \pmod 3\\ 0& n \equiv 0 \pmod 3\end{cases}$$
Usually, when confronted with a sequence, the simplest way to conjecture some closed forms is to use the method of differences.
This is based on the observation that if $a_n = \alpha n^k + \alpha'n^{k'} + \dots$ then we can quickly find the degree $k$ of that polynomial by repeatedly computing differences between successive terms until they're all equal.
Take, for example, take the sequence $-3, -2, 1, 6, 13, 22, 33, \dots$. Computing the differences: $$\begin{matrix} -3 & & -2 & & 1 & & 6 & & 13 & & 22 & & 33 \\ & 1 & & 3 & & 5 & & 7 & & 9 & & 11 & \\ & & 2 & & 2 & & 2 & & 2 & & 2 & & \\ \end{matrix}$$
After $2$ computations the differences are all equal: that means the sequence can be expressed as a polynomial in the form $\alpha n^2 + \beta n + \gamma$. One usually starts from $n^2$ and adjusts accordingly. After some experimentation, you quickly find that in this case $a_n = n^2 - 3$.
Coming to your problems, if you apply this method to your $(1)$ you notice something strange: the method of differences never terminates! After a while you get stuck in a loop and the differences are always the same. This tells us an important bit of information: the sequence cannot be expressed as a polynomial. The simplest thing to try is something exponential. You can also come up with a recursive formula, as @amWhy did, and work from there. From $a_0 = 2, a_n = 3a_{n-1}$ it is obvious then that $a_n = 2 \cdot 3^n$.
For the second one, those repeating digits suggest that $(2)$ has something to do with modulus. In fact, after some experimentation, you can easily come up with something like: $a_n = n^2 \mod 3$.
