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

Question

Aicharya · Accepted Answer

We have, $S_k=\frac{\frac{k-1}{k !}}{1-\frac{1}{k}}=\frac{1}{(k-1) !}$ Now, $ \begin{aligned} & \left(k^2-3 k+1ight) S_k=\{(k-2)(k-1)-1\} \ & =\frac{1}{(k-3) !}-\frac{1}{(k-1) !} \ & \Rightarrow \sum_{k=1}^{100}\left|\left(k^2-3 k+1ight) S_kight| \ & =1+1+2-\left(\frac{1}{99 !}+\frac{1}{98 !}ight)=4-\frac{100^2}{100 !} \end{aligned} $ $ \Rightarrow \quad \frac{100^2}{100 !}+\sum_{k=1}^{100}\left|\left(k^2-3 k+1ight) S_kight|=4 $

Let $S_k, k=1,2, \ldots, 100$, denote the sum of the infinite geometric series whose first term is…

Solution