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?
 A: 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.
A: 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,
$$\boxed{S(x) = e^x(x^3 - 2x^2 + 2x) - 2x}.$$
A: Define $$R(x):=\frac{S(x)}{x}=\sum_{n=0}^{\infty}\frac{x^{n+3}}{(n+3)\cdot n!}$$
Then we have
$$R'(x)=\sum_{n=0}^{\infty} \frac{x^{n+2}}{n!}=x^2 \exp(x)$$
Therefore,
$$R(x)=R(0)+\int_{0}^{x}R'(t)\,dt=\int_{0}^{x}t^2 \exp(t)\,dt$$
It follows that $$\boxed{\:S(x)=x\cdot\int_{0}^{x}t^2 \exp(t)\,dt=\exp(x)(x^3-2x^2+2x)-2x\:\:}$$
A: 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 $\left.\dfrac{S(x)}x\right|_{x\to0}=0$).
