For all n ϵ N, \((1~+~\frac{3}{1})(1~+~\frac{5}{4})(1~+~\frac{7}{9}).......(1~+~(\frac{2n~+~1)}{n^2}))\) is equal to

1
\(\frac{(n~+~1)^2}{2}\)
2
\(\frac{(n~+~1)^3}{3}\)
3
\((n+1)^2\)
4
None of these
5
Not Attempted

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