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JEE Mains · Maths · STD 11 - 8. sequence and series

જો \(a_n=\frac{-2}{4 n^2-16 n+15}\),તો \(a_1+a_2+\ldots \ldots+a_{25}=.........\)

  1. A \(\frac{51}{144}\)
  2. B \(\frac{49}{138}\)
  3. C \(\frac{50}{141}\)
  4. D \(\frac{52}{147}\)
Verified Solution

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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