Let S_k, k=1, 2, ldots, 100,denote the sum of | vedclass

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(D) For $k=1$,the first term is $\frac{1-1}{1!} = 0$,so $S_1 = 0$.For $k \ge 2$,the sum of the infinite geometric series is $S_k = \frac{\frac{k-1}{k!

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$3$

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(D) For $k=1$,the first term is $\frac{1-1}{1!} = 0$,so $S_1 = 0$.For $k \ge 2$,the sum of the infinite geometric series is $S_k = \frac{\frac{k-1}{k!}}{1 - \frac{1}{k}} = \frac{k-1}{k!} 	imes \frac{k}{k-1} = \frac{1}{(k-1)!}$.The expression is $E = \frac{100^2}{100!} + \sum_{k=2}^{100} |(k^2 - 3k + 1) \frac{1}{(k-1)!}|$.Note that $k^2 - 3k + 1 = (k-1)(k-2) - 1$.So,$|(k^2 - 3k + 1) \frac{1}{(k-1)!}| = |\frac{(k-1)(k-2) - 1}{(k-1)!}| = |\frac{k-2}{(k-2)!} - \frac{1}{(k-1)!}|$.Let $f(k) = \frac{k-1}{(k-1)!} = \frac{1}{(k-2)!}$. This is a telescoping sum.Sum $= |\frac{0}{0!} - \frac{1}{1!}| + |\frac{1}{1!} - \frac{2}{2!}| + |\frac{2}{2!} - \frac{3}{3!}| + \dots + |\frac{98}{98!} - \frac{99}{99!}|$.Since $\frac{k-1}{(k-1)!} > \frac{k}{k!}$ for $k \ge 2$,the terms are positive.Sum $= (\frac{0}{0!} - \frac{1}{1!}) + (\frac{1}{1!} - \frac{2}{2!}) + \dots + (\frac{98}{98!} - \frac{99}{99!}) = 0 - \frac{99}{99!} = -\frac{1}{98!}$ (Wait,re-evaluating).Correct telescoping: $\sum_{k=2}^{100} (\frac{k-2}{(k-2)!} - \frac{1}{(k-1)!}) = (0 - 1) + (1 - 1/2) + (2/2 - 1/6) + \dots = 1$.The total value evaluates to $3$.

Let $S_k, k=1, 2, \ldots, 100$,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 $\frac{100^2}{100!} + \sum_{k=1}^{100} |(k^2 - 3k + 1) S_k|$ is

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