Question

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

Answer 1

We have,
$S_k=\frac{\frac{k-1}{k!}}{1-\frac{1}{k}}=\frac{1}{(k-1)!}$
Now, $(k^2-3k+1)s_k=\left\{\right(k-2\left)\right(k-1)-1\}\times \frac{1}{(k-1)!}$
$=\frac{1}{(k-3)!}-\frac{1}{(k-1)!}$
$\Rightarrow {\sum }_{k=1}^{100}\left|\right(k^2-3k+1\left)S_k\right|=1+1+2-(\frac{1}{99!}+\frac{1}{98!})$
$=4-\frac{{100}^2}{100!}$
$\Rightarrow \frac{{100}^2}{100!}+{\sum }_{k=1}^{100}\left|\right(k^2-3k+1\left)S_k\right|=4$