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

माना \(n =1,2, \ldots ., 50\) के लिए, अनन्त गुणोत्तर श्रेणी का योगफल \(S _{ n }\) है जिसका प्रथम पद \(n ^2\) तथा जिसका सार्व अनुपात \(\frac{1}{(n+1)^2}\) है। तब \(\frac{1}{26}+\sum_{ n =1}^{50}\left( S _{ n }+\frac{2}{ n +1}- n -1\right)\) का मान है

  1. A \(41600\)
  2. B \(47651\)
  3. C \(41651\)
  4. D \(41671\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(41651\)

Step-by-step Solution

Detailed explanation

\(S_{n}=\frac{n^{2}}{1-\frac{1}{(n+1)^{2}}}=\frac{n(n+1)^{2}}{(n+2)}\) \(S_{n}=\frac{n\left(n^{2}+2 n+1\right)}{(n+2)}\) \(S_{n}=\frac{n[n(n+2)+1]}{(n+2)}\) \(S_{n}=n\left[n+\frac{1}{n+2}\right]\) \(S_{n}=n^{2}+\frac{n+2-2}{(n+2)}\) \(S_{n}=n^{2}+1-\frac{2}{(n+2)}\) Now…
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