Question

Let $S_k,k=1,2,\dots ,100$, 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 $\frac{100^2}{100!}+\sum _{k=1}^{100}\left|\left(k^2-3k+1\right)S_k\right|$ is _____

Answer 1

Answer: 3

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