Question

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

Answer 1

Answer: 4

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