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?

Your idea is right, just let $x^3$ appear and differentiate
$$\left(\frac{S(x)}x\right)'=\left(\frac{x^3}{3\cdot0!}+\frac{x^4}{4\cdot1!}+\frac{x^5}{5\cdot2!}+\cdots\right)'=\frac{x^2}{0!}+\frac{x^3}{1!}+\frac{x^4}{2!}+\cdots=x^2e^x.$$
Then by integration,
$$\color{green}{\frac{S(x)}x=(x^2-2x+2)e^x-2}$$ (the constant of integration is drawn from $\left.\dfrac{S(x)}x\right|_{x\to0}=0$).
