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

Let for \(n =1,2, \ldots \ldots, 50, S _{ a }\) be the sum of the infinite geometric progression whose first term is \(n ^{2}\) and whose common ratio is \(\frac{1}{(n+1)^{2}}\). Then the value of \(\frac{1}{26}+\sum\limits_{n=1}^{50}\left(S_{n}+\frac{2}{n+1}-n-1\right)\) is equal to

  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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