Later data often show patterns better than early data, so extend your table of partial sums a bit:
$$\begin{array}{rcc}
n:&1&2&3&4&5&6&7&8&9\\
s_n:&1&-1&3&-5&11&-21&43&-85&171
\end{array}$$
Ignoring the signs, it appears that the numbers in the bottom line are approximately doubling each time. Moreover, still ignoring signs, adjacent partial sums add up to a power of $2$: $|s_1|+|s_2|=2^1$, $|s_2|+|s_3|=2^2$, $|s_3|+|s_4|=2^3$, and apparently in general $|s_n|+|s_{n+1}|=2^n$. (If you go back to the definition of the partial sums, you’ll see why this happens.)
If $|s_n|+|s_{n+1}|=2^n$ and $|s_{n+1}|\approx 2|s_n|$, then $3|s_n|\approx 2^n$; this suggests that we should compare $3|s_n|$ with $2^n$:
$$\begin{array}{rcc}
n:&1&2&3&4&5&6&7&8&9\\
s_n:&1&-1&3&-5&11&-21&43&-85&171\\
3|s_n|:&3&3&9&15&33&63&129&255&513\\
2^n:&2&4&8&16&32&64&128&256&512
\end{array}$$
That pattern’s pretty clear: apparently $3|s_n|=2^n+1$ if $n$ is odd, and $3|s_n|=2^n-1$ if $n$ is even. Those cases can be combined as $3|s_n|=2^n+(-1)^{n+1}$, since $(-1)^{n+1}$ is $1$ when $n$ is odd and $-1$ when $n$ is even. And the algebraic sign of $s_n$ appears to be that of $(-1)^{n+1}$, so if these patterns are real,
$$\begin{align*}
s_n&=\frac{(-1)^{n+1}}3\left(2^n+(-1)^{n+1}\right)\\
&=\frac{(-1)^{n+1}2^n}3+\frac{(-1)^{2n+2}}3\\
&=\frac{(-1)^{n+1}2^n+1}3\\
&=\frac{1-(-2)^n}3\;.
\end{align*}$$
This result can then be proved by mathematical induction, but I suspect that you’re not expected to go that far.
If you’ve already learned the summation formula for finite geometric series, you can apply it to get $s_n$ without looking at any patterns at all, and it’s something that you should learn as soon as possible if you don’t already know it. However, skill at pattern-recognition is useful anyway, so I thought that it might be useful to see how the problem can be attacked in that way as well.
Best Answer
$$\sum_{n=1}^\infty \frac{3n-2}{3\cdot 2^n} = \sum_{n=1}^\infty \frac{n}{2^n} - \frac{1}{3}\sum_{n=1}^\infty \frac{1}{2^{n-1}} = -\frac{2}{3} + \sum_{n=1}^\infty \frac{n}{2^n}$$ since both sums converge. To evaluate the latter sum, note that $$\frac{1}{1-x} = \sum_{i=0}^\infty x^i,$$ so that $$\left(\frac{1}{1-x}\right)' = \frac{1}{(1-x)^2} = \sum_{n=1}^\infty nx^{n-1} = 1 + 2x + 3x^2 + \cdots.$$ Then set $x=\frac{1}{2}$, and $$2 = \frac{1/2}{(1-1/2)^2} = \frac{1}{2}\sum_{n=1}^\infty n\left(\frac{1}{2}\right)^{n-1} = \sum_{n=1}^\infty \frac{n}{2^n}.$$ Finally, $$\sum_{n=1}^\infty \frac{3n-2}{3\cdot 2^n} = -\frac{2}{3} + \sum_{n=1}^\infty \frac{n}{2^n} = -\frac{2}{3}+2 = \frac{4}{3}$$ and you are done.