$\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\ldots$ upto $n$ terms $=$

$\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\ldots$ upto $n$ terms $=$
  1. $\frac{1}{4 n+1}$
  2. $\frac{4}{4 n+1}$
  3. $\frac{n}{4 n+1}$
  4. $\frac{4 n+1}{5(4 n+1)}$

Solution

$\begin{aligned} & \text { } T_n=\frac{1}{(4 n-3)(4 n+1)} \\ & =\frac{1}{4}\left\{\frac{1}{4 n-3}-\frac{1}{4 n+1}\right\}, \Sigma \mathrm{T}_n=\frac{1}{4}\left\{1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9} \ldots-\frac{1}{4 n+1}\right\} \\ & =\frac{1}{4}\left\{1-\frac{1}{4 n+1}\right\}=\frac{n}{4 n+1} \end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 1)

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