LTI, causal, discrete time system output
Unless explicitly stated, you shall not assume that the input is zero for $n&lt;0$, hence the convolution sum should be 

$$y(n) = \sum^\infty_{k=-\infty}3^{k+1}u(n-k)=\sum^\infty_{k=0}3^{k+1}u(n-k).$$

In this situation, without making any assumption on $u[n]$ for $n&lt;0$ you can still compute $y[n]$ for $n \geq 0$ based on the recursive relation as you have outlined :

$$y(n) = 3y(n-1)+3u(n) \implies y[3] = 3y[2] + 3u[3] = 201$$ 

What can be said about $u[n]$ for $n&lt;0$ ? Actually nothing but, given that $y[2] = 66$ we may try the folowing :

$$ 
\begin{align}
y[2] &amp;= \sum^2_{k=0}3^{k+1}u(2-k) + \sum^\infty_{k=3}3^{k+1}u(2-k) \\
&amp;= 66 + \sum^\infty_{k=3}3^{k+1}u(2-k) \\
66 &amp;= 66 +   \sum^\infty_{k=3}3^{k+1}u(2-k) \\
0 &amp;= \sum^\infty_{k=3}3^{k+1}u(2-k) 
\end{align}
$$

That the sum in the parenthesis equals zero, has a trivial solution of $x[n] = 0$ for $ n &lt;0$. 

However, we may also ask whether another, nonzero, solution is also possible that the sum be zero? Consider, for example, the input  $$u[n] = (-1)^{2-n} ~~3^{n-3}$$ for $n&lt;0$ which yields for the sum

$$\sum^\infty_{k=3}3^{k+1}u(2-k) = 1-1+1-1+1-...$$

This is an alternating sequence and it does not converge. However, if it's possible to restrict the length of the input to be even for $n&lt;0$, then the sum will be zero too... 

Nevertheless, in order to uniquely specify $u[n]$ for $n&lt;0$ a single sum is not enough. We must know values of output $y[n]$ for all $n&lt;0$ for that. Hence you cannot conlcude that $x[n]=0$ for $n&lt;0$ under the given situation.