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