Evaluating complicated sum Evaluate for a fixed $m\neq 1$   ( $m\in \mathbb{N}$ )
$$\sum _{k=1}^{n}\left[\left( \sum _{i=1}^{k}i^{2}\right) \left(\sum _{k_{1}+k_{2}+...+k_{m}=k}\dfrac {\left( k_{1}+k_{2}+\ldots +k_{m}\right) !} {k_{1}!k_{2}!...k_{m}!}\right)\right]$$

Note that the last part of your summation:
$$\sum_{k_{1}+k_{2}+\cdots+k_{m}=k}{\frac{k!}{k_{1}!k_{2}!\cdots k_{m}!}}=\sum_{k_{1}+k_{2}+\cdots+k_{m}=k}{{k \choose k_{1},k_{2},\cdots,k_{m}}}$$ 
Where ${k \choose k_{1},k_{2},\cdots,k_{m}}$ is the multinomial co-efficient.
Therefore, we can immediately evaluate this, due to the following theorem:
$$\sum_{k_{1}+k_{2}+\cdots+k_{m}=k}{{k \choose k_{1},k_{2},\cdots,k_{m}}}=m^{k}$$
So we now have:
$$\sum_{k=1}^{n}{\left[\left(\frac{k(k
+1)(2k+1)}{6}\right)\left(m^{k}\right)\right]}=\sum_{k=1}^{n}{m^{k}\left(\frac{k}{6}+\frac{k^{2}}{2}+\frac{k^{3}}{3}\right)}=\sum_{k=1}^{n}{\frac{m^{k}k}{6}}+\sum_{k=1}^{n}{\frac{m^{k}k^{2}}{2}}+\sum_{k=1}^{n}{\frac{m^{k}k^{3}}{3}}$$
Which has no nice expansion, but mathematica yields the following:
$$\frac{1}{6 (-1+m)^4}m (6+6 m-6 m^n-6 m^{1+n}-13 m^n n+15 m^{1+n} n-3 m^{2+n} n+m^{3+n} n-9 m^n n^2+21 m^{1+n} n^2-15 m^{2+n} n^2+3 m^{3+n} n^2-2 m^n n^3+6 m^{1+n} n^3-6 m^{2+n} n^3+2 m^{3+n} n^3)$$
Hope this helps!
