Q48 / 84 · JEE Advanced 2010 (Paper 1) · Mathematics
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
Correct answer: 4
Solution
We have, $S_k=\frac{\frac{k-1}{k !}}{1-\frac{1}{k}}=\frac{1}{(k-1) !}$
Now,
$$
\begin{aligned}
& \left(k^2-3 k+1\right) S_k=\{(k-2)(k-1)-1\} \\
& =\frac{1}{(k-3) !}-\frac{1}{(k-1) !} \\
& \Rightarrow \sum_{k=1}^{100}\left|\left(k^2-3 k+1\right) S_k\right| \\
& =1+1+2-\left(\frac{1}{99 !}+\frac{1}{98 !}\right)=4-\frac{100^2}{100 !}
\end{aligned}
$$
$$
\Rightarrow \quad \frac{100^2}{100 !}+\sum_{k=1}^{100}\left|\left(k^2-3 k+1\right) S_k\right|=4
$$
Hence, the correct answer is 4.