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

\(\frac{1}{3^{2}-1}+\frac{1}{5^{2}-1}+\frac{1}{7^{2}-1}+\ldots+\frac{1}{(201)^{2}-1}\) is equal to

  1. A \(\frac{101}{404}\)
  2. B \(\frac{25}{101}\)
  3. C \(\frac{101}{408}\)
  4. D \(\frac{99}{400}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{25}{101}\)

Step-by-step Solution

Detailed explanation

\(T_{n}=\frac{1}{(2 n+1)^{2}-1} \frac{1}{(2 n+2) 2 n}=\frac{1}{4(n)(n+1)}\) \(=\frac{(n+1)-n}{4 n(n+1)}=\frac{1}{4}\left(\frac{1}{n}-\frac{1}{n+1}\right)\) \(S=\frac{1}{4}\left(1-\frac{1}{101}\right)=\frac{1}{4}\left(\frac{100}{101}\right)=\frac{25}{101}\)
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