AP EAMCET · Maths · Sequences and Series
\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\ldots\) upto \(n\) terms \(=\)
- A \(\frac{1}{4 n+1}\)
- B \(\frac{4}{4 n+1}\)
- C \(\frac{n}{4 n+1}\)
- D \(\frac{4 n+1}{5(4 n+1)}\)
Answer & Solution
Correct Answer
(C) \(\frac{n}{4 n+1}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { } T_n=\frac{1}{(4 n-3)(4 n+1)} \\ & =\frac{1}{4}\left\{\frac{1}{4 n-3}-\frac{1}{4 n+1}\right\}, \Sigma \mathrm{T}_n=\frac{1}{4}\left\{1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9} \ldots-\frac{1}{4 n+1}\right\} \\ & =\frac{1}{4}\left\{1-\frac{1}{4…
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