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

योगफल \(\sum_{n=1}^{21} \frac{3}{(4 n-1)(4 n+3)}\) बराबर है

  1. A \(\frac{7}{87}\)
  2. B \(\frac{7}{29}\)
  3. C \(\frac{14}{87}\)
  4. D \(\frac{21}{29}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{7}{29}\)

Step-by-step Solution

Detailed explanation

\(\sum_{n=1}^{21} \frac{3}{(4 n-1)(4 n+3)}=\frac{3}{4} \sum_{n=1}^{21} \frac{1}{4 n-1}-\frac{1}{4 n+3}\) \(=\frac{3}{4}\left[\left(\frac{1}{3}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{11}\right)+\ldots .+\left(\frac{1}{83}-\frac{1}{87}\right)\right]\)…
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