Step 1: Calculate the sum of the infinite geometric series, $S_k$.

The given infinite geometric series has the first term $a = \frac{k-1}{k!}$ and the common ratio $r = \frac{1}{k}$. The sum is denoted by $S_k$.

Case 1: k = 1 For $k=1$, the first term is $a = \frac{1-1}{1!} = 0$. The common ratio is $r = \frac{1}{1} = 1$. Since the first term is 0, the series is $0 + 0 + 0 + \dots$, and its sum is $S_1 = 0$.

Case 2: k > 1 (i.e., k = 2, 3, ..., 100) For these values of $k$, the common ratio $|r| = \frac{1}{k} < 1$. Thus, the sum of the infinite geometric series converges and is given by the formula $S_k = \frac{a}{1-r}$. $S_k = \frac{\frac{k-1}{k!}}{1 - \frac{1}{k}} = \frac{\frac{k-1}{k!}}{\frac{k-1}{k}} = \frac{k-1}{k!} \times \frac{k}{k-1} = \frac{k}{k!}$ Since $k! = k \times (k-1)!$, we have: $S_k = \frac{k}{k(k-1)!} = \frac{1}{(k-1)!}$ So, we have $S_1 = 0$ and $S_k = \frac{1}{(k-1)!}$ for $k \ge 2$.

Step 2: Evaluate the expression inside the summation.

We need to evaluate the sum $\sum\limits_{k = 1}^{100} {\left| {({k^2} - 3k + 1)\,{S_k}} \right|}$. Let's analyze the term inside the absolute value, $T_k = (k^2 - 3k + 1)S_k$.

• For $k=1$: $T_1 = (1^2 - 3(1) + 1)S_1 = (-1)(0) = 0$.
• For $k \ge 2$: $T_k = (k^2 - 3k + 1) \frac{1}{(k-1)!}$.

To handle the absolute value, let's determine the sign of the quadratic factor $k^2 - 3k + 1$. The roots of the equation $x^2 - 3x + 1 = 0$ are $x = \frac{3 \pm \sqrt{9-4}}{2} = \frac{3 \pm \sqrt{5}}{2}$. Numerically, these roots are approximately 0.38 and 2.62. The parabola opens upwards, so the expression is negative between the roots.

• For $k=1$, $1^2 - 3(1) + 1 = -1 < 0$.
• For $k=2$, $2^2 - 3(2) + 1 = 4 - 6 + 1 = -1 < 0$.
• For $k=3$, $3^2 - 3(3) + 1 = 9 - 9 + 1 = 1 > 0$. For all $k \ge 3$, the expression $k^2 - 3k + 1$ is positive.

Step 3: Calculate the summation.

We can split the summation based on the sign of the term: $\sum\limits_{k = 1}^{100} |T_k| = |T_1| + |T_2| + \sum\limits_{k = 3}^{100} |T_k|$

• $|T_1| = |0| = 0$.
• $|T_2| = |(2^2 - 3(2) + 1)S_2| = |(-1) \frac{1}{(2-1)!}| = |-1 \times \frac{1}{1!}| = 1$.
• For $k \ge 3$, since both $(k^2 - 3k + 1)$ and $S_k$ are positive, $|T_k| = T_k$.

The summation becomes: $0 + 1 + \sum\limits_{k = 3}^{100} (k^2 - 3k + 1) \frac{1}{(k-1)!}$ Let's simplify the general term for $k \ge 3$ by rewriting the numerator: $k^2 - 3k + 1 = k^2 - 2k + 1 - k = (k-1)^2 - k$ So, the term is: $\frac{(k-1)^2 - k}{(k-1)!} = \frac{(k-1)^2}{(k-1)!} - \frac{k}{(k-1)!} = \frac{k-1}{(k-2)!} - \frac{k}{(k-1)!}$ This is a telescoping series. Let $a_k = \frac{k}{(k-1)!}$. Then the term is $a_{k-1} - a_k$.

The sum is: $\sum\limits_{k = 3}^{100} (a_{k-1} - a_k) = (a_2 - a_3) + (a_3 - a_4) + \dots + (a_{99} - a_{100}) = a_2 - a_{100}$ Now, we calculate $a_2$ and $a_{100}$: $a_2 = \frac{2}{(2-1)!} = \frac{2}{1!} = 2$ $a_{100} = \frac{100}{(100-1)!} = \frac{100}{99!}$ So, $\sum\limits_{k = 3}^{100} T_k = 2 - \frac{100}{99!}$.

Putting it all together, the value of the full summation is: $\sum\limits_{k = 1}^{100} |T_k| = 1 + \left(2 - \frac{100}{99!}\right) = 3 - \frac{100}{99!}$

Step 4: Calculate the final value of the expression.

The entire expression is: $E = \frac{100^2}{100!} + \sum\limits_{k = 1}^{100} |(k^2 - 3k + 1)S_k|$. First, let's simplify the term $\frac{100^2}{100!}$: $\frac{100^2}{100!} = \frac{100 \times 100}{100 \times 99!} = \frac{100}{99!}$ Now, substitute the results from the previous steps: $E = \frac{100}{99!} + \left(3 - \frac{100}{99!}\right) = 3$

The final value of the expression is 3.