jmc

More generally, let $$S_n=\sum _{k=1}^n(-1{)}^k\cdot \frac{k^2+k+1}{k!}$$for a positive integer $n.$ We can compute the first few values of $S_n$: \renewcommand{\arraystretch}{1.5} \begin{array}{c|c} n & S_n \\ \hline 1 & -3 \\ 2 & \frac{1}{2} \\ 3 & -\frac{5}{3} \\ 4 & -\frac{19}{24} \\ 5 & -\frac{21}{20} \\ 6 & -\frac{713}{720} \end{array} \renewcommand{\arraystretch}{1}First, the denominators seem to be factors of $n!.$ Second, the fractions seem to be getting close to $-1.$ So, we re-write each sum in the form $\frac{\ast }{n!}-1$: \renewcommand{\arraystretch}{1.5} \begin{array}{c|c} n & S_n \\ \hline 1 & \frac{-2}{1!} - 1 \\ 2 & \frac{3}{2!} - 1 \\ 3 & \frac{-4}{3!} - 1 \\ 4 & \frac{5}{4!} - 1 \\ 5 & \frac{-6}{5!} - 1 \\ 6 & \frac{7}{6!} - 1 \\ \end{array} \renewcommand{\arraystretch}{1}Now the pattern is very clear: It appears that $$S_n=(-1{)}^n\cdot \frac{n+1}{n!}-1.$$So, set $T_n=(-1{)}^n\cdot \frac{n+1}{n!}-1.$ Since we expect the sum to telescope, we can compute the difference $T_k-T_{k-1}$: $$\begin{array}{rl}{T}_k-T_{k-1}& =(-1{)}^k\cdot \frac{k+1}{k!}-1-(-1{)}^{k-1}\cdot \frac{k}{(k-1)!}+1\\ & =(-1{)}^k\cdot \frac{k+1}{k!}+(-1{)}^k\cdot \frac{k}{(k-1)!}\\ & =(-1{)}^k\cdot \frac{k+1}{k!}+(-1{)}^k\cdot \frac{k^2}{k!}\\ & =(-1{)}^k\cdot \frac{k^2+k+1}{k!}.\end{array}$$Thus, indeed the sum telescopes, which verifies our formula $$S_n=(-1{)}^n\cdot \frac{n+1}{n!}-1.$$In particular, $$S_{100}=\frac{101}{100!}-1.$$Then $a=101,$ $b=100,$ and $c=1,$ so $a+b+c=\boxed{202}.$