For any n ∈ N, \(\frac{1}{2.5}+\frac{1}{5.8}+\ldots+\frac{1}{(3 n−1)(3 n+2)}\) =

1
\(\frac{n}{6n+4}\)
2
\(\frac{n^{2}}{6n+4}\)
3
\(\frac{1}{2}.\frac{n^{2}}{6n+4}\)
4
\(\frac{n}{6n^{2}+4}\)

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