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JEE Mains · Maths · STD 11 - 8. sequence and series
If \([\mathrm{x}]\) be the greatest integer less than or equal to \(\mathrm{x}\), then \(\sum_{\mathrm{n}=8}^{100}\left[\frac{(-1)^{n} \mathrm{n}}{2}\right]\) is equal to:
- A \(-2\)
- B \(4\)
- C \(2\)
- D \(0\)
Answer & Solution
Correct Answer
(B) \(4\)
Step-by-step Solution
Detailed explanation
\(\sum_{n=8}^{100}\left[\frac{(-1)^{n} \cdot n}{2}\right]=\left[\frac{8}{2}\right]+\left[\frac{-9}{2}\right]+\left[\frac{10}{2}\right]+\left[\frac{-11}{2}\right]+\ldots+\ldots\left[\frac{-99}{2}\right]+\left[\frac{100}{2}\right]\) \(=4-5+5-6+6+\ldots-50+50=4\)
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