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?

I'm assuming you've made a typo and you actually have
$$S(x) = \frac{x^4}{3(0)!} + \frac{x^5}{4(1)!}+\frac{x^{{\color{red}6}}}{5(2)!}+ \cdots.$$
Consider $P(x) = S(x)/x$. We have
\begin{align}
P(x) &= \frac{x^3}{3(0)!} + \frac{x^4}{4(1)!}+ \dfrac{x^5}{5(2)!} + \cdots\\
&=\int_0^x\left(\dfrac{t^2}{0!} + \dfrac{t^3}{1!} + \dfrac{t^4}{2!} + \cdots\right){\rm d}t\\
&= \int_0^x t^2e^t{\rm d}t\\
&= e^x(x^2 - 2x + 2) - 2.
\end{align}
Thus,
$$\boxed{S(x) = e^x(x^3 - 2x^2 + 2x) - 2x}.$$
