This is hard 2

simplify as: $\begin{array}{c}=&\dfrac{n^2+n+1}{(n^2+n)*(n+1)!}\\ =&\dfrac{(n+1)^2}{n*(n+1)^2*n!}-\dfrac{n}{n*(n+1)*(n+1)!}\\ =&\dfrac{1}{n*n!}-\dfrac{1}{(n+1)*(n+1)!} \end{array}$ hence $\sum_{n=1}^{2011} (\dfrac{n^2+n+1}{(n^2+n)*(n+1)!})=\sum_{n=1}^{2011} (\dfrac{1}{n*n!})-\sum_{n=1}^{2011}(\dfrac{1}{(n+1)*(n+1)!})=\sum_{n=1}^{2011} (\dfrac{1}{n*n!})-\sum_{n=2}^{2012}(\dfrac{1}{n*n!})$ we see that all terms except last and first cancels out,hence $\sum_{n=1}^{2011} (\dfrac{n^2+n+1}{(n^2+n)*(n+1)!})=\dfrac{1}{1*1!}-\dfrac{1}{2012*2012!}=\dfrac{2012*2012!-1}{2012*2012!}$ comparing the question $a!-b!=2012*2012!$ note that $(n+1)!-n!=n!(n+1-1)=n*n!$ hence $2013!-2012!=2012*2012!$ and $2012+2013=\boxed{4025}$

i think it should be in level 3

$\frac{n^2+n+1}{(n^2+n)(n+1)}=\frac{1}{n \times n!}-\frac{1}{(n+1) \times (n+1)!}$ $\Sigma^{2011}_{n=1}\frac{n^2+n+1}{(n^2+n)(n+1)}=\frac{2013!-2012!-1}{2013!-2012!}$ $a+b=\boxed{4025}$