Gifting summations

$\dfrac{3^{r}(r)!(3r^2+5r+1)}{r^2+3r+2} = \dfrac{3^{r+1}(r+1)!}{r+2} - \dfrac{3^{r}(r)!}{r+1}$

Now enjoy telescoping :)

Let $\displaystyle S_k = \sum_{r=0}^k \frac {3^rr!(3r^2+5r+1)}{r^2+3r+2}$ . Then we can prove by induction that $S_k = \dfrac {3^{k+1}(k+1)!}{k+2}-1$ for $k=0$ .

Proof:

For $k=0$ , $S_0 = \dfrac {3^00!(1)}{2} = \dfrac 12 = \dfrac {3^{0+1}(0+1)!}{0+2}-1$ . Therefore, the claim is true for $k=0$ . Now assuming the claim is true for $k$ , then we have:

$\begin{aligned} S_{k+1} & = S_k + \frac {3^{k+1}(k+1)!(3(k+1)^2+5(k+1)+1)}{(k+1)^2+3(k+1)+2} \\ & = \frac {3^{k+1}(k+1)!}{k+2}-1 + \frac {3^{k+1}(k+1)!(3k^2+11k+9)}{(k+2)(k+3)} \\ & = \frac {3^{k+1}(k+1)!}{k+2} \left(1+\frac {3k^2+11k+9}{k+3}\right) - 1 \\ & = \frac {3^{k+1}(k+1)!}{k+2} \left(\frac {3k^2+12k+12}{k+3}\right) - 1 \\ & = \frac {3^{k+1}(k+1)!}{k+2} \left(\frac {3(k+2)^2}{k+3}\right) - 1 \\ & = \frac {3^{k+2}(k+2)!}{k+3} - 1 \end{aligned}$

This implies that the claim is also true for $k+1$ and hence true for all $k \ge 0$ .

Therefore, $S_{10} = \dfrac {3^{11}11!}{12} - 1 = \dfrac {3^{11}11!-12}{12}$ , $\implies m-n = 11 - 11 = \boxed{0}$ .