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

${\text{S}}_{\text{k}}=\frac{\frac{\text{k}-1}{\text{k}!}}{1-\frac{1}{\text{k}}}=\frac{1}{(\text{k}-1)!},\text{ for k}>1$

$$\begin{gathered} \sum _{\text{k}=2}^{100}\mid ({\text{k}}^2-3\text{k}+1)\frac{1}{(\text{k}-1)!}\mid \\ =\sum _{\text{k}=2}^{100}\left|\frac{{(\text{k}-1)}^2-\text{k}}{(\text{k}-1)!}\right| \\ =\sum _{\text{k}=2}^{100}|\frac{\text{k}-1}{(\text{k}-2)!}-\frac{\text{k}}{(\text{k}-1)!}| \\ =|\frac{1}{0!}-\frac{2}{1!}|+|\frac{2}{1!}-\frac{3}{2!}|+|\frac{3}{2!}-\frac{4}{3!}|+\cdots \\ =\frac{2}{1!}-\frac{1}{0!}+\frac{2}{1!}-\frac{3}{2!}+\frac{3}{2!}-\frac{4}{3!}+\cdots +\frac{99}{98!}-\frac{100}{99!} \\ =1+\frac{2}{1!}-\frac{100}{99!} \\ =3-\frac{100}{99!} \end{gathered}$$

Hence the value of $\frac{{100}^2}{100!}+\sum _{\text{k}=1}^{100}\left|({\text{k}}^2-3\text{k}+1){\text{S}}_{\text{k}}\right|$

$\begin{array}{l}=\frac{100}{99!}+3-\frac{100}{99!}\\ =3\end{array}$