Mathematical Induction problem

[mateomy]

Im practicing inductions on my own and I am getting stuck on one in particular...

$$S_n = \frac{3(3^n-1)}{2} for a_n = 3^n$$

I know that
$$S_1 = 3$$

when you plug 1 into the equation, because it is the first term in the sequence
Therefore,
$$S_1 = a_1 = 3$$

I need to prove then
$$S_{n+1} = \frac{3(3^{n+1} -1)}{2}$$

I know that I need to add S*sub(n) to a*sub(n+1)
so doing that I get
$$\frac{3(3^n - 1) + 2(3^{n+1})}{2}$$

I don't understand how to get it to match with what I am supposed to prove.

Did I go in the right steps?

 

[lurflurf]

It might help to write
s_n=(a_{n+1}+a_1)/2
then
s_{n+1}=s_n+a_{n+1}
=(a_{n+1}+a_1)/2+a_{n+1}
then endeavor to show
s_{n+1}=(a_{n+2}+a_1)/2

hint
a_{n+1}=3 a_n