<s>[INST] Evaluating complicated sum Evaluate for a fixed $m\neq 1$   ( $m\in \mathbb{N}$ )
$$\sum _{k=1}^{n}\left[\left( \sum _{i=1}^{k}i^{2}\right) \left(\sum _{k_{1}+k_{2}+...+k_{m}=k}\dfrac {\left( k_{1}+k_{2}+\ldots +k_{m}\right) !} {k_{1}!k_{2}!...k_{m}!}\right)\right]$$
 [/INST] As Shaktal showed earlier it is remains to find closed form for
$$
\sum\limits_{k=1}^n\frac{m^k k^3}{3}
$$
Consider the following equality
$$
\sum\limits_{k=1}^n x^k=\frac{x(x^n-1)}{x-1}
$$
After triple differentiation we get
$$
\sum\limits_{k=1}^n kx^{k-1}=\frac{d}{dx}\frac{x(x^n-1)}{x-1}\\
\sum\limits_{k=1}^n k(k-1)x^{k-2}=\frac{d^2}{dx^2}\frac{x(x^n-1)}{x-1}\\
\sum\limits_{k=1}^n k(k-1)(k-2)x^{k-3}=\frac{d^3}{dx^3}\frac{x(x^n-1)}{x-1}
$$
Now let's make a small trick
$$
\sum\limits_{k=1}^n\frac{k^3 x^k}{3}=
\frac{1}{3}\sum\limits_{k=1}^n (k (k-1)(k-2)+3k(k-1)+k)x^k=
\frac{1}{3}\left(x^3\sum\limits_{k=1}^n k (k-1)(k-2)x^{k-3} +3x^2\sum\limits_{k=1}^n k(k-1)x^{k-2} +x\sum\limits_{k=1}^n k x^{k-1} \right)=
\frac{1}{3}\left(x^3\frac{d^3}{dx^3}\frac{x(x^n-1)}{x-1} +3x^2\frac{d^2}{dx^2}\frac{x(x^n-1)}{x-1} +x\frac{d}{dx}\frac{x(x^n-1)}{x-1} \right)
$$
It is remains to simplify this expression (i.e. differentiante) and then substitute $x=m$
</s>