JEE Mains · Maths · STD 11 - 8. sequence and series
If \(a_n=\frac{-2}{4 n^2-16 n+15}\), then \(a_1+a_2+\ldots \ldots+a_{25}\) is equal to :
- A \(\frac{51}{144}\)
- B \(\frac{49}{138}\)
- C \(\frac{50}{141}\)
- D \(\frac{52}{147}\)
Answer & Solution
Correct Answer
(C) \(\frac{50}{141}\)
Step-by-step Solution
Detailed explanation
If \(a_n=\frac{-2}{4 n^2-16 n+15}\) then \(a_1+a_2+\ldots \ldots \ldots a_{25}\) \(\sum \limits_{n=1}^{25} a_n=\sum \frac{-2}{4 n^2-16 n+15}\) \(=\sum \frac{-2}{4 n^2-6 n-10 n+15}\) \(=\sum \frac{-2}{2 n(2 n-3)-5(2 n-3)}\) \(=\sum \frac{-2}{(2 n-3)(2 n-5)}\)…
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