JEE Advanced · Mathematics · 5. Sequences & Series
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
- A 2
- B 4
- C 6
- D 8
Answer & Solution
Correct Answer
(B) 4
Step-by-step Solution
Detailed explanation
We have, \(S_k=\frac{\frac{k-1}{k !}}{1-\frac{1}{k}}=\frac{1}{(k-1) !}\)
Now,
\(\left(k^2-3 k+1\right) S_k=\{(k-2)(k-1)-1\} \)
\( =\frac{1}{(k-3) !}-\frac{1}{(k-1) !} \)
\( \Rightarrow \sum_{k=1}^{100}\left|\left(k^2-3 k+1\right) S_k\right| \)
\( =1+1+2-\left(\frac{1}{99 !}+\frac{1}{98 !}\right)=4-\frac{100^2}{100 !}\)
\(\Rightarrow \frac{100^2}{100 !}+\sum_{k=1}^{100}\left|\left(k^2-3 k+1\right) S_k\right|=4\)
Now,
\(\left(k^2-3 k+1\right) S_k=\{(k-2)(k-1)-1\} \)
\( =\frac{1}{(k-3) !}-\frac{1}{(k-1) !} \)
\( \Rightarrow \sum_{k=1}^{100}\left|\left(k^2-3 k+1\right) S_k\right| \)
\( =1+1+2-\left(\frac{1}{99 !}+\frac{1}{98 !}\right)=4-\frac{100^2}{100 !}\)
\(\Rightarrow \frac{100^2}{100 !}+\sum_{k=1}^{100}\left|\left(k^2-3 k+1\right) S_k\right|=4\)
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