Let $S_{k}, k=1,2, \ldots ., 100$, denote the sum of the infinite geometric series whose first term

Question

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

Professor PC Thomas Classes · Answer

(3) : $S_{k}=\frac{k-1}{\left\{k\left(1-\frac{1}{k}ight)ight\}}=\frac{1}{k-1}, k \geq 2$ For $k=1, S_{1}=0$\ $\sum_{k=1}^{100}\left|\left(k^{2}-3 k+1ight) \frac{1}{|k-1|}ight|$\ $=0+\sum_{k=1}^{100}\left|\left(k^{2}-3 k+1ight) \frac{1}{|k-1|}ight|$\ $\Rightarrow \quad \sum_{k=2}^{100}\left|\frac{k-1}{\mid k-2}-\frac{k}{|k-1|}ight|$\ $=1+\sum_{k=3}^{100}\left|\frac{k-1}{\mid k-2}-\frac{k}{|k-1|}ight|$\ $=1+\left(\frac{2}{\lfloor 1}-\frac{3}{\lfloor 2}ight)+\left(\frac{3}{\lfloor 2}-\frac{4}{\underline{3}}ight)+\ldots\left(\frac{99}{\underline{98}}-\frac{100}{\underline{99}}ight)$\ $=1+2-\frac{100}{\underline{99}}$\ Thus, $\frac{100^{2}}{\lfloor 100}+\sum_{k=1}^{100}\left|\left(k^{2}-3 k+1ight) S_{k}ight|=3$\