Induction proof on a sequence

Couple of things –

1. $S(k)$ must have a truth value. In this case it's the statement "$a_k = \frac{2^{k+1} + (-1)^k}{3}$". So you shouldn't write $S(k+1) = S(k)+2S(k-1)$ because that's adding statements!
2. The question asks to show $S(k)$ for $k\geq0$. So your base case must include $k=0$. (Note that here I've abbreviated "$S(k)$ is true" to "$S(k)$" – in the same way that I can abbreivate "it's true that it's raining" to "it's raining". But feel free to say "$S(k)$ is true" for clarity.)
3. The aim for the induction part is to (as I think you know) assume $S(k-1)$ and $S(k)$, and then show $S(k+1)$. The aim is then to do something like $$ \begin{eqnarray} a_{k+1} &=& a_k + 2a_{k-1} &\text{(by definition)}\\ &=& \frac{2^{k+1} + (-1)^k}{3} + \frac{2^{k} + (-1)^{k-1}}{3} &\text{(by inductive hypothesis)} \\ &=& \dots &\text{(some steps for you to work out)} \\ &=& \frac{2^{k+2} + (-1)^{(k+1)}}{3} \end{eqnarray} $$ i.e. $S(k+1)$. In your proof you seemed to begin with assuming $S(k+1)$, which is incorrect.