What is the solution of this summation? $$S(x) = \frac{x^4}{3(0)!} + \frac{x^5}{4(1)!}+\frac{x^6}{5(2)!}+.....$$
If the first term was $$x^3$$ and the next terms were $$x^{3+i}$$ then differentiating it would have given $$x^2.e^x$$ and then it was possible to integrate it. But how to solve this one?

What you have is
$$S(x) = \sum_{k=0}^\infty \frac{x^{k+4}}{(k+3) \cdot k!}$$
Divide the inside by $x$ and multiply the outside by it too:
$$S(x) = x \sum_{k=0}^\infty \frac{x^{k+3}}{(k+3) \cdot k!}$$
Now, take the derivative, assuming you can do so termwise (a similar argument to that for $e^x$ might work). Then you get
$$S'(x) = \sum_{k=0}^\infty \frac{x^{k+3}}{(k+3) \cdot k!} + x \sum_{k=0}^\infty \frac{x^{k+2}}{k!}$$
The first sum is just $S(x)/x$, no big deal there. The second can clearly be seem to be $x^3e^x$ through a quick factorization. Thus, you get the ODE
$$S'(x) = \frac{S(x)}{x} + x^3 e^x$$
What would some initial conditions be? Clearly, $S(0) = S'(0) = 0$, so let's use those.

This ODE seems fairly simple to solve, so I'll leave the remainder of the calculations up to you.
