Every finite-dimensional complex representation of $sl_n(\mathbb C)$ (or every other semisimple complex Lie algebra) is completely reducible, i.e. decomposes as sum of irreducible representations. To classify those irreducible representations ("irreps") is one of the main points of any book, article, blog post or set of lecture notes on this, of which there are plenty.
Note that one always has the $1$-dimensional trivial irrep.
The results are kind of easy for $sl_2$, where for each dimension $k$, there is exactly one irrep; for $k=1$, this is the trivial one, for $k=2$ the defining ("standard") and fundamental one, for $k=3$ the adjoint one, then it goes higher.
For $sl_3$, it's already more involved, as the irreps can be classified by pairs $(p,q) \in \mathbb Z^{\ge 0}$, where the irrep indexed by $(p,q)$ has dimension $\frac12 (p+1)(q+1)(p+q+2)$. So we have
- one $1$-dimensional irrep, the trivial one, index $(0,0)$;
- two $3$-dimensional irreps, both fundamental (in math terminology), one of them the defining ("standard") one and the other its dual, indices $(0,1)$ and $(1,0)$;
- two $6$-dimensional irreps (thanks @Nate Stemen for the comment), indexed by $(2,0)$ and $(0,2)$, respectively;
- one $8$-dimensional irrep, the adjoint one, index $(1,1)$;
- then it goes higher.
So if you want to see a representation of $sl_3$ on $\mathbb C^k$ with dimension $k=4$ or $5$ (or $6$ or $7$), you cannot have an irreducible one. For $k=4$ you can have a sum of the trivial one and either of the $3$-dimensional ones; or four trivial ones. For $k=5$ you have a sum of either of the $3$-dimensional ones and two copies of the trivial one; or five trivial ones.
For $k=6$, there are those two new irreps, indexed by $(2,0)$ and $(0,2)$, respectively. Alternatively, you could add the two different $3$-dimensional ones. Or choose one $3$-dimensional one and add it to itself. Or choose one $3$-dimensional one and add three trivial ones. I think you can see how it goes from there. Starting at $k=8$ you'll have another building block, so plenty of options for $k=9, 10, ...$ etc.
For $sl_4$, the first irreps occur in dimension $k=1$ (trivial), $k=4$ (two: the standard one, and its dual), $k=6$ (another fundamental one, in math terminology), and it goes higher.
One of the greatest results is that all these building blocks occur in some tensor product of the fundamental ones. So when you complain in your comment
"for $sl(3,C)$ he does lot of examples but all in the standard $V=C^3$ or its dual or their tensor product spaces. for $sl(4,C)$ he does the same again as in he works on vector space $C^4$ , its dual and their several inner product spaces."
you are touching on (but slightly missing) the coolest result: Namely, that by doing that, "he" (the authors) are actually talking about all representations. Namely, all irreps are contained as summands in what they do there. They leave the easy part of putting together those irreps to match some dimension $k$ you like to you (or me, as outlined above).
