$\frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + \dots$ to n terms =

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$\frac{n}{2 n + 3}$

none of these

$\frac{n}{3 (2 n + 3)}$

$\frac{1}{(n + 2) (n + 4)}$

answer is A.

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Detailed Solution

For $n = 1,$ have

$\frac{n}{3 (2 n + 3)} = \frac{1}{35} =$ First term of the series

For $n = 2,$ we have

$\frac{n}{3 (2 n + 3)} = \frac{2}{3.7} = \frac{1}{35} + \frac{1}{5.7} =$ Sum of first two terms of the series

Hence, option (a) is correct

$A L I T E R$ We have

$\frac{1}{35} + \frac{1}{5.7} + \frac{1}{7.9} +$ to $n$ terms

$\begin{array}{l} = \frac{1}{2} [\frac{2}{35} + \frac{2}{5.7} + \frac{2}{79} + \dots + \frac{2}{(2 n + 1) (2 n + 3)}] \\ = \frac{1}{2} [(\frac{1}{3} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{7}) + (\frac{1}{7} - \frac{1}{9}) + \dots + (\frac{1}{2 n + 1} - \frac{1}{2 n + 3})] \\ = \frac{1}{2} [\frac{1}{3} + \frac{1}{2 n + 3}] = \frac{n}{3 (2 n + 3)} \end{array}$