sequences and series - Value of this sum?

Let $S_k$, $k=1,2,3,4...,100$ denote the sum of infinite geometric series whose first term is $\frac{k-1}{k!}$ and the common ratio is $\frac{1}{k}$. Then the value of $$\frac{100^2}{100!}+\sum^{100}_{k=1}|(k^2-3k+1)S_k| \text{ is?}$$

All I could figure out in this one is:

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

But I'm not sure on how to continue. Please help

Answer

What you found for $S_k$ simplifies to $S_k={1\over(k-1)!}$. Now $k^2-3k+1\gt0$ if $k\ge3$, so you are asking for the value of

$${100^2\over100!}+S_1+S_2+\sum_{k=3}^{100}{k^2-3k+1\over(k-1)!}={100\over99!}+1+1+\sum_{k=3}^{100}{(k-1)^2-k\over(k-1)!}$$

But

$$\sum_{k=3}^{100}{(k-1)^2-k\over(k-1)!}=\sum_{k=3}^{100}{k-1\over(k-2)!}-\sum_{k=3}^{100}{k\over(k-1)!}=\sum_{k=2}^{99}{k\over(k-1)!}-\sum_{k=3}^{100}{k\over(k-1)!}={2\over1}-{100\over99!}$$

and thus

$${100^2\over100!}+\sum_{k=1}^{100}|(k^2-3k+1)S_k|=4$$