JEE Mains · Maths · STD 11 - 8. sequence and series
If \(\frac{1}{2 \times 3 \times 4}+\frac{1}{3 \times 4 \times 5}+\frac{1}{4 \times 5 \times 6}+\ldots+\) \(\frac{1}{100 \times 101 \times 102}=\frac{ k }{101}\), then \(34\,k\) is equal to \(.....\)
- A \(285\)
- B \(284\)
- C \(286\)
- D \(283\)
Answer & Solution
Correct Answer
(C) \(286\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{2.3 .4}+\frac{1}{3.4 .5}+\ldots .++\frac{1}{100.101 .102}=\frac{ k }{101}\) \(\frac{4-2}{2.3 .4}+\frac{5-3}{3.4 .5}+\ldots . .+\frac{102-100}{100.101 .102}=\frac{2 k }{101}\)…
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