As has been said before, induction works as follows. Let $S_k$ be a statement parameterised with a natural number $k$. If you can show that $S_1$ holds, and that $S_k$ implies $S_{k + 1}$ for any $k$ (actually, you may assume that it holds for all $n \leq k$, but for simple problems like these the weaker form is usually sufficient) then by induction you will have shown $S_k$ holds for all natural $k$.
Here your statement $S_k$ is:
$$S_k ~ : ~ 3 + 3^2 + \cdots + 3^k = \frac{1}{2} \left ( 3^{k + 1} - 3 \right )$$
So we first begin by showing it is true for $S_1$, i.e. with $k = 1$, and indeed:
$$\frac{1}{2} \left ( 3^2 - 3 \right ) = \frac{6}{2} = 3$$
Okay, part 1 done. Now we ASSUME that $k$ is any integer, and that the statement $S_k$ is true. In other words, we ASSUME that the following is true (it is given to us):
$$3 + 3^2 + \cdots + 3^k = \frac{1}{2} \left ( 3^{k + 1} - 3 \right )$$
We need to show that IF THIS IS TRUE, THEN $S_{k + 1}$ MUST ALSO BE TRUE, i.e. we must show that:
$$3 + 3^2 + \cdots + 3^k = \frac{1}{2} \left ( 3^{k + 1} - 3 \right ) ~ ~ ~ \stackrel{IMPLIES}{\implies} ~ ~ 3 + 3^2 + \cdots + 3^k + 3^{k + 1} = \frac{1}{2} \left ( 3^{(k + 1) + 1} - 3 \right )$$
We can show this as follows. First, start from our assumption:
$$3 + 3^2 + \cdots + 3^k = \frac{1}{2} \left ( 3^{k + 1} - 3 \right )$$
Now add $3^{k + 1}$ to each side (remember that if $x = y$, then $x + c = y + c$, i.e. $x = y$ implies $x + c = y + c$, and in fact the reverse is also true in this case, i.e. the two statements are equivalent - but for induction we do not need equivalence, only implication). We then get:
$$3 + 3^2 + \cdots + 3^k + 3^{k + 1} = \frac{1}{2} \left ( 3^{k + 1} - 3 \right ) + 3^{k + 1}$$
At this point the LHS is equal to the LHS of the statement we want to arrive at ($S_{k + 1}$), so we really just need to show that:
$$\frac{1}{2} \left ( 3^{k + 1} - 3 \right ) + 3^{k + 1} = \frac{1}{2} \left ( 3^{(k + 1) + 1} - 3 \right )$$
Once you have proven this, then you will have shown that $S_k$ implies $S_{k + 1}$, for any $k$ (we made no assumptions on the value of $k$ here, it could be any natural number). Therefore it follows logically that:
$$S_1 \implies S_2 \implies S_3 \implies \cdots$$
And we also know $S_1$ is true (we showed it earlier). Therefore $S_k$ is true for all natural numbers $k$. In order words, the equation holds for all natural numbers, and the proof by induction is complete.
Does this make it clearer? Do you understand the logical steps taken and why induction works? Can you complete the missing part necessary to show that $S_k \implies S_{k + 1}$ (the "inductive step")?