Let   denote the sum of the infinite geometric series whose first term is   and the common ratio is  . Then the value of $\frac{{100}^2}{100!}+\sum _{\mathrm{k}=1}^{100} \left({\mathrm{k}}^2-3\mathrm{k}+1\right){\mathrm{S}}_{\mathrm{K}}$  is____ 

$\begin{array}{l}{\mathrm{S}}_{\mathrm{k}}=\frac{\frac{\mathrm{k}-1}{\mathrm{k}!}}{1-\frac{1}{\mathrm{k}}}=\frac{1}{(\mathrm{k}-1)!},\text{ for }\mathrm{k}>1\\ \sum _{\mathrm{k}=2}^{100} \left|\left({\mathrm{k}}^2-3\mathrm{k}+1\right)\frac{1}{(\mathrm{k}-1)!}\right|\\ =\sum _{\mathrm{k}=2}^{100} \left|\frac{(\mathrm{k}-1{)}^2-\mathrm{k}}{(\mathrm{k}-1)!}\right|\end{array}$
$\begin{array}{l}=\sum _{\mathrm{k}=2}^{100} \left|\frac{\mathrm{k}-1}{(\mathrm{k}-2)!}-\frac{\mathrm{k}}{(\mathrm{k}-1)!}\right|\\ =\left|\frac{1}{0!}-\frac{2}{1!}\right|+\left|\frac{2}{1!}-\frac{3}{2!}\right|+\left|\frac{3}{2!}-\frac{4}{3!}\right|+\cdots \\ =\frac{2}{1!}-\frac{1}{0!}+\frac{2}{1!}-\frac{3}{2!}+\frac{3}{2!}-\frac{4}{3!}+\cdots +\frac{99}{98!}-\frac{100}{99!}\\ =1+\frac{2}{1!}-\frac{100}{99!}\\ =3-\frac{100}{99!}\end{array}$
$$\begin{gathered} \frac{{100}^2}{100!}+\sum _{k=1}^{100} \left|\left(k^2-3k+1\right)S_k\right| \\ \begin{array}{l}=\frac{100}{99!}+3-\frac{100}{99!}\\ =3\end{array} \end{gathered}$$