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

$S_k=\frac{\frac{k-1}{k!}}{1-\frac{1}{k}}=\frac{1}{(k-1)!}$

$\left(k^2-3k+1\right)S_k=\frac{(\mathrm{k}-1{)}^2-\mathrm{k}}{(\mathrm{k}-1)!}$

$\sum _{\mathrm{k}=2}^{100} \left|\frac{(\mathrm{k}-1{)}^2-\mathrm{k}}{(\mathrm{k}-1)!}\right|=\sum _{\mathrm{k}=2}^{100} \left|\frac{\mathrm{k}-1}{(\mathrm{k}-2)!}-\frac{\mathrm{k}}{(\mathrm{k}-1)!}\right|$

$=\left|\frac{1}{0!}-\frac{2}{1!}\right|+\left|\frac{2}{1!}-\frac{3}{2!}\right|+\left|\frac{3}{2!}-\frac{4}{3!}\right|+\cdots$

$=\frac{2}{1!}-\frac{1}{0!}+\frac{2}{1!}-\frac{3}{2!}+\frac{3}{2!}-\frac{4}{3!}+\cdots +\frac{99}{98!}-\frac{100}{99!}$

$=1+\frac{2}{1!}-\frac{100}{99!}$

$=3-\frac{100}{99!}$

$\Rightarrow \frac{{100}^2}{100!}+\sum _{k=1}^{100} \left|\left(k^2-3k+1\right)S_k\right|=4$